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Page No 474:
Question 1:
In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
(i)
(ii)
(iii) tan θ = 11
(iv)
Answer:
(i) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Perpendicular side = 2 and
Hypotenuse = 3
Therefore, by Pythagoras theorem,
Now we substitute the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)
Therefore,
Hence, Base =
Now,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(ii) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Base = 4 and
Hypotenuse = 5
Therefore,
By Pythagoras theorem,
Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)
Hence, Perpendicular side = 3
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(iii) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Base = 1 and
Perpendicular side = 5
Therefore,
By Pythagoras theorem,
Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC)
Hence, Hypotenuse =
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(iv) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Base = 12 and
Perpendicular side = 5
Therefore,
By Pythagoras theorem,
Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC)
Hence, Hypotenuse = 13
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(x) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Base = 5 and
Hypotenuse = 13
Therefore,
By Pythagoras theorem,
Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)
Hence, Perpendicular side = 12
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(xi) Given:
…… (1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Perpendicular side = 1 and
Hypotenuse =
Therefore,
By Pythagoras theorem,
Now we substitute the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)
Hence, Base side = 3
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
(xii) Given: ……(1)
By definition,
…... (2)
By Comparing (1) and (2)
We get,
Base = 12 and
Hypotenuse = 15
Therefore,
By Pythagoras theorem,
Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)
Hence, Perpendicular side = 9
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Now,
Therefore,
Page No 474:
Question 2:
In a âABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine
(i) sin A, cos A
(ii) sin C, cos C
Answer:
(i) The given triangle is below:-
Given: In ,
To Find:
In this problem, Hypotenuse side is unknown
Hence we first find Hypotenuse side by Pythagoras theorem
By Pythagoras theorem,
We get,
By definition,
By definition,
Answer:
(ii) The given triangle is below:
Given: In ΔABC,
To Find:
In this problem, Hypotenuse side is unknown
Hence we first find Hypotenuse side by Pythagoras theorem
By Pythagoras theorem,
We get,
By definition,
By definition,
Answer:
Page No 474:
Question 3:
Given 15 cot A = 8, find sin A and sec A.
Answer:
Given: 15 = 8
To Find:
Since
By taking 15 on R.H.S
We get,
By definition,
Hence,
Comparing equation (1) and (2)
We get,
= 8
= 15
can be drawn as shown below using above information
Hypotenuse side AC is unknown.
Therefore, we find side AC of by Pythagoras theorem.
So, by applying Pythagoras theorem to
We get,
Therefore, Hypotenuse = 17
Now by definition,
Substituting values of sides from the above figure
By definition,
Hence,
Answer: and
Page No 474:
Question 4:
If , evaluate :
(i)
(ii) cot2 θ
Answer:
(i) Given:
To evaluate:
…… (1)
We know the following formula
By applying the above formula in the numerator of equation (1) ,
We get,
… (Where a = 1 and b = )
…… (2)
Similarly,
By applying formula in the denominator of equation (1) ,
We get,
… (Where a = 1 and b = )
…… (3)
Substituting the value of numerator and denominator of equation (1) ,from equation (2) and(3)
Therefore,
…… (4)
Since,
Therefore,
Putting the value of and in Equation (4)
We get,
We know that,
Since,
Therefore,
Answer:
(ii) Given:
To evaluate:
Squaring on both sides,
We get,
Answer:
Page No 474:
Question 5:
If 3 cot A = 4, check whether or not.
Answer:
Given:
To check whether or not
Dividing by 3 on both sides,
We get,
…… (1)
By definition,
Therefore,
…… (2)
Comparing Equation (1) and (2)
We get,
= 4
= 3
Hence, is as shown in figure below
In, Hypotenuse is unknown
Hence, It can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem in
We get
Hence, Hypotenuse = 5
To check whether or not
We get the values of
By definition,
Substituting the value from Equation (1)
We get,
….… (3)
Now by definition,
…… (4)
Now by definition,
…… (5)
Now we first take L.H.S of Equation
Substituting value of from equation (3)
We get,
Taking L.C.M on both numerator and denominator
We get,
…… (6)
Now we take R.H.S of Equation
Substituting value of and from equation (4) and (5)
We get,
…… (7)
Comparing Equation (6) and (7)
We get,
Answer: Yes
Page No 474:
Question 6:
If , find the value of .
Answer:
Given:
…… (1)
Now, we know that
Therefore equation (1) becomes as follows
Now, by applying invertendo
We get,
Now, by applying Compenendo-dividendo
We get,
Therefore,
Page No 474:
Question 7:
If 3 cot θ = 2, find the value of .
Answer:
Given:
Therefore,
…… (1)
Now, we know that
Therefore equation (1) becomes
…… (2)
Now, by applying Invertendo to equation (2)
We get,
…… (3)
Now, multiplying by on both sides
We get,
Therefore, 3 cancels out on R.H.S and
We get,
Now by applying dividendo in above equation
We get,
…… (4)
Now, multiplying by on both sides of equation (3)
We get,
Therefore, 2 cancels out on R.H.S and
We get,
Now by applying componendo in above equation
We get,
…… (5)
Now, by dividing equation (4) by equation (5)
We get,
Therefore,
Therefore, on L.H.S cancels out and we get,
Now, by taking 2 in the numerator of L.H.S on the R.H.S
We get,
Therefore, 2 cancels out on R.H.S. and
We get,
Hence,
Page No 474:
Question 8:
If , prove that .
Answer:
Given:
…… (1)
Now, we know that
Therefore equation (1) becomes
…… (2)
Now, multiplying by on both sides of equation (2)
We get,
Therefore,
…... (3)
Now by applying dividendo in above equation (3)
We get,
…… (4)
Now by applying componendo in equation (3)
We get,
…… (5)
Now, by dividing equation (4) by equation (5)
We get,
Therefore,
Therefore, and cancels on L.H.S and R.H.S respectively and we get,
Hence, it is proved that
Page No 474:
Question 9:
If , show that .
Answer:
Given:
To show that
Now, we know that
Therefore,
Therefore,
…… (1)
Now, we know that
…… (2)
Now, by comparing equation (1) and (2)
We get,
= 5
And
Hypotenuse = 13
Therefore from above figure
Base side
Hypotenuse
Side AB is unknown, It can be determined by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
Therefore by substituting the values of known sides
We get,
Therefore,
Therefore,
…… (3)
Now, we know that
Now from figure (a)
We get,
Therefore,
…… (4)
Now L.H.S. of the equation to be proved is as follows
Substituting the value of and from equation (1) and (4) respectively
We get,
Therefore,
Hence proved that,
Page No 474:
Question 10:
If , show that sin θ (1 − tan θ).
Answer:
Given: ……(1)
To show that
Now, we know that …… (2)
Therefore, by comparing equation (1) and (2)
We get,
= 12
And
Hypotenuse = 13
Therefore from above figure
Base side
Hypotenuse
Side AB is unknown and it can be determined by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
Therefore by substituting the values of known sides
We get,
Therefore,
Therefore,
…… (3)
Now, we know that
Now from figure (a)
We get,
Therefore,
…… (4)
Now, we know that
Now from figure (a)
We get,
Therefore,
…… (5)
Now L.H.S. of the equation to be proved is as follows
…… (6)
Substituting the value of and from equation (4) and (5) respectively
We get,
Taking L.C.M inside the bracket
We get,
Therefore,
Now, by opening the bracket and simplifying
We get,
…… (7)
From equation (6) and (7) , it can be shown that
Page No 474:
Question 11:
If , show that
Answer:
Given: ……(1)
To show that
Now, we know that
Since ……(2)
Therefore,
Comparing Equation (1) and (2)
We get,
Therefore, Triangle representing angle is as shown below
Hypotenuse AC is unknown and it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
Therefore by substituting the values of known sides
We get,
Therefore,
Therefore,
…… (3)
Now, we know that
Now from figure (a)
We get,
Therefore from figure (a) and equation (3) ,
…… (4)
Now, we know that
Therefore, from equation (4)
We get,
Therefore,
…… (5)
Now, we know that
Now from figure (a)
We get,
Therefore from figure (a) and equation (3) ,
…… (6)
Now, we know that
Therefore, from equation (6)
We get,
Therefore,
…… (7)
Now, L.H.S of the equation to be proved is as follows
Substituting the value of and from equation (6) and (7)
We get,
Now by taking L.C.M. in numerator as well as denominator
We get,
Therefore,
Therefore,
Therefore,
Hence proved that
Page No 474:
Question 12:
If , find the value of .
Answer:
Given: ……(1)
To find the value of
Now we know that
Therefore,
Therefore from equation (1)
…… (2)
Also, we know that
Therefore,
Substituting the value of from equation (2)
We get,
Therefore
…… (3)
Also, we know that .
Therefore,
Therefore
Therefore,
Therefore,
…… (4)
Also
Therefore, from equation (4)
We get,
…… (5)
Substituting the value of,,and from equation (2) (3) (4) and (5) respectively in the expression below
We get,
Therefore,
Page No 474:
Question 13:
If , find the value of .
Answer:
Given: ……(1)
To find the value of
Now, we know the following trigonometric identity
Therefore, by substituting the value of from equation (1) ,
We get,
Therefore,
Therefore by taking square root on both sides
We get,
Therefore,
…… (2)
Now, we know that
Therefore by substituting the value of and from equation (2) and (1) respectively
We get,
…… (4)
Now, by substituting the value of and from equation (2) and (4) respectively in the expression below
We get,
Therefore,
Therefore,
Page No 474:
Question 14:
In the given figure, find tan P and cot R. Is tan P = cot R?
Answer:
The given figure is below:
To Find:
In the given right angled ΔPQR, length of side QR is unknown.
Therefore, to find length of side QR we use Pythagoras Theorem
Hence, by applying Pythagoras theorem in ΔPQR,
We get,
Now, we substitute the length of given side PR and PQ in the above equation
By definition, we know that
… (1)
Also, by definition, we know that
… (2)
Comparing equation (1) and (2) , we come to know that R.H.S of both the equation are equal
Therefore, L.H.S of both the equation are also equal
Answer:
Page No 475:
Question 15:
If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Answer:
Given:
…… (1)
To show:
is as shown in figure below
Now since ……from (1)
Therefore
Now observe that denominator of above equality is same that is AB
Hence only when
Therefore …… (2)
We know that when two sides of a triangle are equal, then angle opposite to the sides are also equal.
Therefore from equation (2)
We can say that
Angle opposite to side AC = Angle opposite to side BC
Therefore,
Hence,
Page No 475:
Question 16:
In a âABC, right angled at A, if , find the value of sin B cos C + cos B sin C.
Answer:
Given:
To find:
The givenis as shown in figure below
Side BC is unknown and can be found using Pythagoras theorem
Therefore,
Now by substituting the value of known sides from figure (a)
We get,
Now by taking square root on both sides
We get,
Therefore Hypotenuse side BC = 2 …… (1)
Now
Therefore,
Now by substituting the values from equation (1) and figure (a)
We get,
…… (2)
Now
Therefore,
Now by substituting the values from equation (1) and figure (a)
We get,
…… (3)
Now
Therefore,
Now by substituting the values from equation (1) and figure (a)
We get,
…… (4)
Now by definition,
Therefore,
Now by substituting the value of and from equation (4) and given data respectively
We get,
Now gets cancelled as it is present in both numerator and denominator
Therefore,
…… (5)
Now by substituting the value of and from equation (2) , (3) , (4) and (5) respectively in
We get,
Page No 475:
Question 17:
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) for some angle θ.
Answer:
(i) In, is acute an angle
Therefore,
Minimum value of is 0° and
Maximum value of is 90°
We know that and
tan90° = ∞
Therefore the statement that;
“The value of is always less than1” is false
(ii)
In and, is acute angle
Therefore,
Minimum value of is 0°and
Maximum value of is
We know that cos0° = 1 and
cos90° = 0
Now,
Therefore minimum value of is …… (1)
Now,
Therefore maximum value of is …… (2)
Now consider the given value
Here,
This value 2.4 lies in between 1 and
Now from equation (1) and (2) , we can say that the value lies in between minimum value of (that is 1) and maximum value of (that is)
Hence, , for some value of angle A is true
(iii) Cosecant of angle A is defined as
Also, is defined as
Therefore,
…… (1)
And
is defined as …… (2)
Therefore from equation (1) and (2) , it is clear that and (that is cosecant of angle A) are two different trigonometric angles
Hence, is the abbreviation used for cosecant of angle A is False
(iv) cot A is a trigonometric ratio which means cotangent of angle A
Hence, is the product of cot and A is False
(v)
The value
In, is acute an angle
Therefore,
Minimum value of is 0° and
Maximum value of is 90°
We know that and
sin90° = 1
Therefore the value of should lie between 0 and 1 and must not exceed 1
Hence the given value for (that is) is not possible
Therefore, , for some angle = False
Page No 475:
Question 18:
If , find the value of .
Answer:
Given: ……(1)
To Find: The value of expression
Now, we know that
…… (2)
Now when we compare equation (1) and (2)
We get,
= 12
And
Hypotenuse = 13
Therefore, Triangle representing angle is as shown below
Base side BC is unknown and it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
Therefore by substituting the values of known sides
We get,
Therefore,
Therefore,
…… (3)
Now, we know that
Now from figure (a)
We get,
Therefore from figure (a) and equation (3) ,
…… (4)
Now we know that,
Therefore, substituting the value of and from equation (1) and (4)
We get,
Therefore 13 gets cancelled and we get
…… (5)
Now we substitute the value of , and from equation (1) , (4) and (5) respectively in the expression below
Therefore,
We get,
Therefore by further simplifying we get,
Now 169 gets cancelled and gets reduced to
Therefore
Therefore the value of is
That is
Page No 475:
Question 19:
If , verify that .
Answer:
Given:
…… (1)
To verify:
…… (2)
Now we know that
Therefore
Now, by substituting the value of from equation (1)
We get,
Therefore,
…… (3)
Now, we know the following trigonometric identity
Therefore,
Now by substituting the value of from equation (3)
We get,
Now by taking L.C.M
We get,
Now, by taking square root on both sides
We get,
Therefore,
…… (4)
Now, we know that
Now by substituting the value of and from equation (3) and (4) respectively
We get,
Therefore
…… (5)
Now from the expression of equation (2)
Now by substituting the value of and from equation (3) and (4)
We get,
Therefore,
Now by taking L.C.M of both numerator and denominator
We get,
…… (6)
Now from the expression of equation (2)
Now by substituting the value of from equation (5)
We get,
Now by taking L.C.M
We get,
Now,
Therefore,
Therefore,
…… (7)
Now by comparing equation (6) and (7)
We get,
Page No 475:
Question 20:
If , prove that .
Answer:
Given:
…… (1)
To prove:
Now we know is defined as follows
…… (2)
Now by comparing equation (1) and (2)
We get,
= 3
= 4
Therefore triangle representing angle is as shown below
Side AC is unknown and can be found using Pythagoras theorem
Therefore,
Now by substituting the value of known sides from figure
We get,
Now by taking square root on both sides
We get,
Therefore Hypotenuse side AC = 5 …… (3)
Now we know, is defined as follows
Therefore from figure (a) and equation (3)
We get,
…… (4)
Now we know
Therefore by substituting the value of from equation (4)
We get,
Therefore,
…… (5)
Now we know, is defined as follows
Therefore from figure (a) and equation (3)
We get,
…… (6)
Now we know
Therefore by substituting the value of from equation (6)
We get,
Therefore,
…… (7)
Now, in expression, by substituting the value of andfrom equation (6) and (7) respectively, we get,
L.C.M of 3 and 4 is 12
Now by taking L.C.M in above expression
We get,
Now 12 gets cancelled and we get,
Now
Therefore,
Now 5 gets cancelled and we get,
Therefore, it is proved that
Page No 475:
Question 21:
If 3 cos θ − 4 sin θ = 2 cos θ + sin θ, find tan θ.
Answer:
Given:
To find:
Now consider the given expression
Now by dividing both sides of the above expression by
We get,
Now by separating the denominator for each terms
We get,
Now in the above expression present in both numerator and denominator gets cancelled
Therefore,
…… (1)
Now we know that,
Therefore by substituting in equation (1)
We get,
Now by taking on L.H.S
We get,
Therefore,
Hence
Page No 485:
Question 1:
Evaluate each of the following
sin 45° sin 30° + cos 45° cos 30°
Answer:
We have,
…… (1)
Now
So by substituting above values in equation (1)
We get,
Therefore,
Page No 485:
Question 2:
Evaluate each of the following
cos 60° cos 45° − sin 60° sin 45°
Answer:
We have to find the value of the following expression
…… (1)
Now,,
So by substituting above values in equation (1)
We get,
Therefore,
Page No 485:
Question 3:
Evaluate each of the following
sin2 30° cos2 45° + 4 tan2 30° +
Answer:
We have,
…… (1)
Now,
,, ,,
So by substituting above values in equation (1)
We get,
LCM of 8, 3, 2 and 24 is 48
Therefore by taking LCM
We get,
In the above equation the first term gets reduced to
Therefore,
Page No 485:
Question 4:
Evaluate each of the following
4 (sin460° + cos4 30°) − 3 (tan2 60° − tan2 45°) + 5 cos2 45°
Answer:
We have,
…… (1)
Now,
,, ,
So by substituting above values in equation (1)
We get,
Now, gets reduced to
Therefore,
Now, gets reduced to
Therefore,
Now by taking LCM
We get,
Therefore,
Page No 486:
Question 5:
Evaluate each of the following
(cosec2 45° sec2 30°) (sin2 30° + 4 cot2 45° − sec2 60°)
Answer:
We have,
…… (1)
Now,
, , , ,
So by substituting above values in equation (1)
We get,
Now, in above equation 4 cancel 8 and 2 remains
Hence,
Therefore,
Page No 486:
Question 6:
Evaluate each of the following
cosec3 30° cos 60° tan3 45° sin2 90° sec2 45° cot 30°
Answer:
We have,
…… (1)
Now,
, , , , ,
So by substituting above values in equation (1)
We get,
Now, 2 gets cancelled and we get,
Page No 486:
Question 7:
Evaluate each of the following
Answer:
We have,
…… (1)
Now,
,, ,
So by substituting above values in equation (1)
We get,
Now, in the third term 4 gets cancelled by 2 and 2 remains
Therefore,
Now in the second term, 4 gets cancelled by 2 and 2 remains
Therefore,
Now, LCM of denominator in the above expression is 6
Therefore by taking LCM
We get,
Now in the above expression, gets reduced to
Therefore,
Page No 486:
Question 8:
Evaluate each of the following
Answer:
We have,
…… (1)
Now,
, , , , ,
So by substituting above values in equation (1)
We get,
Now,
3 gets cancel in numerator and we get,
Now, in the numerator get reduced to 2and we get,
Therefore,
Page No 486:
Question 9:
Evaluate each of the following
Answer:
We have,
…… (1)
Now,
,,
So by substituting above values in equation (1)
We get,
Now by taking terms with denominator 2 together and solving
We get,
Now gets reduced to -2
Therefore,
Therefore,
Page No 486:
Question 10:
Find the value of x in each of the following :
(i)
(ii)
(iii)
Answer:
(i) We have,
Since,
Now,
(ii) We have,
Now by cross multiplying we get,
…… (1)
Now we know that
…… (2)
Therefore from equation (1) and (2)
We get,
…… (3)
Since,
…… (4)
By comparing equation (3) and (4) we get,
(iii) We have,
…… (1)
Now we know that
and
Now by substituting above values in equation (1), we get,
Therefore,
…… (2)
Since,
…… (3)
Therefore by comparing equation (2) and (3)
We get,
Page No 486:
Question 11:
If θ = 30°, verify that
(i)
(ii)
Answer:
(i) Given:
…… (1)
To verify:
…… (2)
Now consider right hand side
Hence it is verified that,
(ii) Given:
…… (1)
To verify:
…… (2)
Now consider left hand side of the equation (2)
Therefore,
Now consider right hand side of equation (2)
Therefore,
Hence it is verified that,
Page No 486:
Question 12:
If A = B = 60°, verify that
(i) cos (A − B) = cos A cos B + sin A sin B
(ii) sin (A − B) = sin A cos B − cos A sin B
(iii)
Answer:
(i) Given:
…… (1)
To verify:
…… (2)
Now consider left hand side of the expression to be verified in equation (2)
Therefore,
Now consider right hand side of the expression to be verified in equation (2)
Therefore,
Hence it is verified that,
(ii) Given:
…… (1)
To verify:
…… (2)
Now consider LHS of the expression to be verified in equation (2)
Therefore,
Now by substituting the value of A and B from equation (1) in the above expression
We get,
Hence it is verified that,
(iii) Given:
…… (1)
To verify:
…… (2)
Now consider LHS of the expression to be verified in equation (2)
Therefore,
Now consider RHS of the expression to be verified in equation (2)
Therefore,
Now by substituting the value of A and B from equation (1) in the above expression
We get,
Hence it is verified that,
Page No 486:
Question 13:
Prove that .
Answer:
Consider the left hand side.
Now, consider the right hand side.
So, LHS = RHS
Hence proved.
Page No 486:
Question 14:
(i) If find A and B.
(ii) If , then find the values of A and B.
Answer:
(i)
Given:
…… (1)
…… (2)
We know that,
…… (3)
…… (4)
Now by comparing equation (1) and (3)
We get,
…… (5)
Now by comparing equation (2) and (4)
We get,
…… (6)
Now to get the values of A and B, let us solve equation (5) and (6) simultaneously
Therefore by adding equation (5) and (6)
We get,
Therefore,
Hence
Now by subtracting equation (5) from equation (6)
We get,
Therefore,
Hence
Therefore the values of A and B are as follows
and
(ii)
Also,
Adding (1) and (2), we get
Putting in (1), we get
Thus, the values of A and B are 37.5º and 7.5º, respectively.
Page No 486:
Question 15:
If find A and B.
Answer:
Given:
…… (1)
…… (2)
We know that,
…… (3)
…… (4)
Now by comparing equation (1) and (3)
We get,
…… (5)
Now by comparing equation (2) and (4)
We get,
…… (6)
Now to get the values of A and B, let us solve equation (5) and (6) simultaneously
Therefore by adding equation (5) and (6)
We get,
Therefore,
Hence
Now by subtracting equation (5) from equation (6)
We get,
Therefore,
Hence
Therefore the values of A and B are as follows
and
Page No 486:
Question 16:
In a âABC right angled at B, ∠A = ∠C. Find the values of
(i) sin A cos C + cos A sin C
(ii) sin A sin B + cos A cos B
Answer:
(i) We have drawn the following figure related to given information
To find:
…… (1)
Now we have,
,
,
Now by substituting the above values in equation (1)
We get,
Therefore,
…… (2)
Now in right angled
By applying Pythagoras theorem
We get,
Now, by substituting above value of AC2 in equation (2)
We get,
Now both numerator and denominator contains
Therefore it gets cancelled and 1 remains
Hence
(ii) We have drawn the following figure
e
To find:
…… (1)
Now we know that sum of all the angles of any triangle is 180°
Therefore,
Since and
Therefore,
It is given that
Therefore,
…… (2)
Now we have,
,
,
Now by substituting the above values in equation (1)
We get,
Since
Therefore
Page No 486:
Question 17:
Find acute angles A and B, if .
Answer:
Given:
…… (1)
…… (2)
We know that,
…… (3)
…… (4)
Now by comparing equation (1) and (3)
We get,
…… (5)
Now by comparing equation (2) and (4)
We get,
…… (6)
Now to get the values of A and B, let us solve equation (5) and (6) simultaneously
Therefore by subtracting equation (5) from (6)
We get,
Therefore,
Hence
Now by multiplying equation (5) by 2
We get,
…… (7)
Now by subtracting equation (6) from (7)
We get,
Therefore,
Hence
Therefore the values of A and B are as follows and
Page No 486:
Question 18:
In âPQR, right-angled at Q, PQ = 3 cm and PR = 6 cm. Determine ∠P and ∠R.
Answer:
We are given the following information in the form of triangle
To find: and
Now, in
…… (1)
Now we know that
…… (2)
Now by comparing equation (1) and (2)
We get,
…… (3)
Now we have
Now we know that
Therefore,
Now by cross multiplying
We get,
Therefore,
cm …… (4)
Now we know that
Now we know,
…… (6)
Now by comparing equation (5) and (6)
We get,
…… (7)
Hence from equation (3) and (7)
and
Page No 486:
Question 19:
If sin (A − B) = sin A cos B − cos A sin B and cos (A − B) = cos A cos B + sin A sin B, find the values of sin 15° and cos 15°.
Answer:
Given:
…… (1)
…… (2)
To find:
The values of and
In this problem we need to find and
Hence to get angle we need to choose the value of A and B such that
So If we choose and
Then we get,
Therefore by substituting and in equation (1)
We get,
Therefore,
…… (3)
Now we know that,
, ,
Now by substituting above values in equation (3)
We get,
Therefore,
…… (4)
Now by substituting and in equation (2)
We get,
Therefore,
…… (5)
Now we know that,
, ,
Now by substituting above values in equation (5)
We get,
Therefore,
…… (6)
Therefore from equation (4) and (6)
Page No 486:
Question 20:
In a right triangle ABC, right angled at C, if ∠B = 60° and AB = 15 units. Find the remaining angles and sides.
Answer:
We are given the following triangle with related information
It is required to find , and length of sides AC and BC
is right angled at C
Therefore,
Now we know that sum of all the angles of any triangle is
Therefore,
…… (1)
Now by substituting the values of known angles and in equation (1)
We get,
Therefore,
Therefore,
Now,
We know that,
Now we have,
AB=15 units and
Therefore by substituting above values in equation (2)
We get,
Now by cross multiplying we get,
Therefore,
…… (3)
Now,
We know that,
Now we have,
AB=15 units and
Therefore by substituting above values in equation (4)
We get,
Now by cross multiplying we get,
Therefore,
Hence,
Page No 486:
Question 21:
If âABC is a right triangle such that ∠C = 90°, ∠A = 45° and BC = 7 units. Find ∠B, AB and AC.
Answer:
We are given the following information in the form of the triangle
It is required to find and length of sides AB and AC
In
Now we know that sum of all the angles of any triangle is
Therefore,
…… (1)
Now by substituting the values of known angles and in equation (1)
We get,
Therefore,
Therefore,
…… (2)
Now,
We know that,
Now we have,
BC = 7 units and
Therefore by substituting above values in equation (3)
We get,
Now by cross multiplying we get,
Therefore,
…… (4)
Now,
We know that,
……(5)
Now we have,
and
Therefore by substituting above values in equation (5)
We get,
Now by cross multiplying we get,
Therefore,
…… (6)
Therefore,
From equation (2), (4) and (6)
, ,
Page No 486:
Question 22:
If A and B are acute angles such that , find A + B.
Answer:
Given:
…… (1)
…… (2)
…… (3)
Now by substituting the value of and from equation (1) and (2) in equation (3)
We get,
Therefore,
…… (3)
Now we know that
…… (4)
Now by comparing equation (3) and (4)
We get,
Page No 492:
Question 1:
Evaluate the following :
(i)
(ii)
(iii)
Answer:
(i) Given that
Since
Therefore
(ii) Given that
Since
Therefore
(iii) Given that
Since
Page No 492:
Question 2:
Evaluate the following :
(i)
(ii)
(iii)
(iv)
(v) tan 48° tan 23° tan 42° tan 67°
(vi) sec50° sin 40° + cos40° cosec 50°
Answer:
(i) We have to find:
Since and
So
So the value of is .
(ii) We have to find:
Since and
So the value of is .
(v) We have to find:
Since and
So the value of is .
(iv) We have to find:
Sinceand
So
So the value of is .
(v) We have to find
Since.So
So the value of is .
(vi) We find to find
Since, and .So
So the value of is .
Page No 492:
Question 3:
Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°
(i) cos 78° + sec 78°
(ii) cosec 54° + sin 72°
(iii) cot 85° + cos 75°
(iv) sin 67° + cos 75°
Answer:
(i) We knowand
Thus the desired expression is .
(ii) We know and.So
Thus the desired expression is .
(iii) We know thatand.So
Thus the desired expression is.
(iv) We know thatand.So
Thus the desired expression is .
Page No 492:
Question 4:
Express cos 75° + cot 75° in terms of angles between 0° and 30°.
Answer:
Given that:
Hence the correct answer is
Page No 492:
Question 5:
If sin 3A = cos (A − 26°), where 3A is an acute angles, find the value of A.
Answer:
We are given 3A is an acute angle
We have:
Hence the correct answer is
Page No 492:
Question 6:
If A, B, C are the interior angles of a triangle ABC, prove that
(i)
(ii)
Answer:
(i) We have to prove:
Since we know that in triangle
Proved
(ii) We have to prove:
Since we know that in triangle
Proved
Page No 492:
Question 7:
Prove the following :
(i) sin θ sin (90° − θ) − cos θ cos (90° − θ) = 0
(ii)
(iii)
(iv)
(v) sin (50° − θ) − cos (40° − θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° = 1
Answer:
(i) We have to prove:
Left hand side
=Right hand side
Proved
(ii) We have to prove:
Left hand side
= right hand side
Proved
(iii) We have to prove:
Left hand side
= right hand side
Proved
(iv) We have to prove:
Left hand side
= Right hand side
Proved
(v) We have to prove:
Left hand side
Since .So
=Right hand side
Proved
Page No 492:
Question 8:
Evaluate :
(i) tan 7° tan 23° tan 60° tan 67° tan 83°
(ii)
(iii)
(iv)
Answer:
We have to evaluate the following values-
(i) We will use the properties of complementary angles.
(ii) We will use the properties of complementary angles.
(iii) We will use the properties of complementary angles.
(xi)
Page No 493:
Question 9:
If sin θ = cos (θ − 45°), where θ and θ − 45° are acute angles, find the degree measure of θ.
Answer:
Given that: where and are acute angles
We have to find
Therefore
Page No 493:
Question 10:
If A, B, C are the interior angles of a âABC, show that :
(i)
(ii)
Answer:
(i) We have to prove:
Since we know that in triangle
Dividing by 2 on both sides, we get
Proved
(ii) We have to prove:
Since we know that in triangle
Dividing by 2 on both sides, we get
Proved
Page No 493:
Question 11:
If sin 3 θ = cos (θ − 6°), where 3 θ and θ − 6° are acute angles, find the value of θ.
Answer:
We have: where and are acute angles
We have to find
Now we proceed as to find
Therefore
Page No 493:
Question 12:
If sec 4A = cosec (A − 20°), where 4A is an acute angles, find the value of A.
Answer:
Given: and is an acute angle
We have to find
Now
Hence the value of is
Page No 493:
Question 13:
If sec 2A = cosec (A − 42°), where 2A is an acute angles, find the value of A.
Answer:
Given: and is an acute angle
We have to find
So we proceed as follows to calculate
Hence the value of is
Page No 493:
Question 14:
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Answer:
Given that,
tan2A = cot(A− 18°)
cot(90° − 2A) = cot(A − 18°) (âµ tanθ = cot(90° − θ))
90° − 2A = A − 18°
108° = 3A
A = 36°Page No 493:
Question 15:
Prove that: sin (50° + θ) − cos (40° − θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° = 1
Answer:
We have to prove: sin (50° + θ) − cos (40° − θ) + tan 1° tan 10° tan 20° tan 70° tan 80° tan 89° = 1
Left hand side
= 1 × 1 × 1 []
= Right hand side
Proved.
Page No 493:
Question 16:
Evaluate :
(i)
(ii)
(iii)
Answer:
We have to evaluate the following values-
(i) We will use the values of known angles of different trigonometric ratios.
(ii) We will use the properties of complementary angles.
(iii) We will use the properties of complementary angles.
Page No 493:
Question 17:
If 2θ + 45° and 30° − θ are acute angles, find the degree measure of θ satisfying sin (2θ + 45°) = cos (30° − θ).
Answer:
Given that: where and are acute angles
We have to find
So we have
Hence the value of is
Page No 493:
Question 1:
Write the maximum and minimum values of sin θ.
Answer:
The maximum value of is and the minimum value of is because value of lies between −1 and 1
Page No 493:
Question 2:
Write the maximum and minimum values of cos θ.
Answer:
The maximum value of is and the minimum value of is because value of lies between −1 and 1
Page No 493:
Question 3:
What is the maximum value of ?
Answer:
The maximum value of is because the maximum value of is that is
Page No 493:
Question 4:
What is the maximum value of ?
Answer:
The maximum value of is because the maximum value of is that is
Page No 493:
Question 5:
If , find the value of .
Answer:
It is given that .
We have to find .
Page No 494:
Question 6:
If , find the value of .
Answer:
Given in question:
We have to find
Hence the value of is
Page No 494:
Question 7:
If 3 cot θ = 4, find the value of .
Answer:
We have:
Since we know that in right angle triangle
Now, we find
Hence the value of is
Page No 494:
Question 8:
Given , what is the value of ?
Answer:
Given: ,
We know that:
Now we find,
Hence the value of is
Page No 494:
Question 9:
If , write the value of .
Answer:
Given:
Now we find,
Hence the value of is
Page No 494:
Question 10:
If , then what is the value of cot B?
Answer:
Given that:
Hence the value of is
Page No 494:
Question 11:
If A + B = 90° and , what is the value of sin A?
Answer:
We have:
Hence the value of is
Page No 494:
Question 12:
Write the acute angle θ satisfying .
Answer:
We have:
Hence the acute angle is
Page No 494:
Question 13:
Write the value of cos 1° cos 2° cos 3° ....... cos 179° cos 180°.
Answer:
Given that:
Hence the value of is
Page No 494:
Question 14:
Write the value of tan 10° tan 15° tan 75° tan 80°?
Answer:
We have to find:
Hence the value of is
Page No 494:
Question 15:
If A + B = 90° and , what is cot B?
Answer:
Given in question:
Hence the value of is
Page No 494:
Question 16:
Find A, if tan 2A = cot(A – 24°).
Answer:
Thus, the value of A is 38º.
Page No 494:
Question 17:
Find the value of sin233° + sin257°.
Answer:
We have
Thus, the value of sin233° + sin257° is 1.
Page No 494:
Question 18:
Evaluate sin2 60° + 2tan 45° – cos2 30°.
Answer:
Page No 494:
Question 19:
If sin , calculate sec A.
Answer:
We know
Now,
Page No 494:
Question 20:
If , find the value of (sin A + cos A) sec A.
Answer:
Given:
We know that:
Now we find,
Hence the value of is
Page No 494:
Question 21:
In the given figure, PS = 3 cm, QS = 4 cm, ∠PRQ = θ, ∠PSQ = 90°, PQ ⊥ RQ and RQ = 9 cm. Evaluate tan θ.
Answer:
In right âPQS,
In right âPQR,
Thus, the value of tanθ is .
Page No 494:
Question 1:
The value of (sin30° + cos30°) – (sin60° + cos60°) is ________.
Answer:
The value of (sin30° + cos30°) – (sin60° + cos60°) is ___0___.
Page No 494:
Question 2:
The value of is ______.
Answer:
The value of is ___1___.
Page No 495:
Question 3:
The value of (sin 45° + cos 45°)3 is _______.
Answer:
The value of (sin 45° + cos 45°)3 is .
Page No 495:
Question 4:
The value of (sin 30° + cos 30°)2 – (sin 60° – cos 60°)2 is __________.
Answer:
The value of (sin 30° + cos 30°)2 – (sin 60° – cos 60°)2 is .
Page No 495:
Question 5:
The value of (cos2 23° – sin2 67°) is ____________.
Answer:
We know that,
The value of (cos2 23° – sin2 67°) is ______0______.
Page No 495:
Question 6:
The value of the expression is ________.
Answer:
We know that,
The value of the expression is ____2____.
Page No 495:
Question 7:
Given that sin α = and cos β = , then α + β = _________.
Answer:
Also,
Given that sin α = and cos β = , then α + β = ____90°____.
Page No 495:
Question 8:
Given that sin (α – β ) = and cos (α + β) = , then α = _______ β = _________.
Answer:
Also,
Adding (1) and (2), we get
Putting in (2), we get
Given that sin (α – β ) = and cos (α + β) = , then α = ___3___ β = ____45º___.
Page No 495:
Question 9:
If 4 tan θ = 3, then is equal to _________.
Answer:
(Dividing numerator and denominator by cosθ)
If 4 tan θ = 3, then is equal to .
Page No 495:
Question 10:
If cos 5θ = sinθ and 5θ < 90°, then the value of tan 3θ is ___________.
Answer:
If cos 5θ = sinθ and 5θ < 90°, then the value of tan 3θ is ____1____.
Page No 495:
Question 11:
The value of sec2 60° – tan2 60° is __________.
Answer:
The value of sec2 60° – tan2 60° is _____1_____.
Page No 495:
Question 12:
If A + B = 90°, then the value of tan2A – cot2B is _________.
Answer:
If A + B = 90°, then the value of tan2A – cot2B is ____0____.
Page No 495:
Question 13:
The value of cos 1° cos 2° cos 3° ...... cos 120° is _______.
Answer:
The value of cos 1° cos 2° cos 3° ...... cos 120° is ___0___.
Page No 495:
Question 14:
If tan θ + cot θ = 2 and, 0° < θ < 90° then tan10 θ + cot10 θ is equal to _________.
Answer:
So,
If tan θ + cot θ = 2 and, 0° < θ < 90° then tan10 θ + cot10 θ is equal to ____2____.
Page No 495:
Question 15:
If sin θ + cos θ = 1 and 0° ≤ θ ≤ 90°, then the possible values of θ are ________.
Answer:
The given equation is sinθ + cosθ = 1.
When θ = 0°,
LHS = sinθ + cosθ = sin0° + cos0° = 0 + 1 = 1 = RHS
When θ = 90°,
LHS = sinθ + cosθ = sin90° + cos90° = 1 + 0 = 1 = RHS
Thus, the possible values of θ (0° ≤ θ ≤ 90°) satisfying the given equation are 0° and 90°.
If sin θ + cos θ = 1 and 0° ≤ θ ≤ 90°, then the possible values of θ are _0° and 90°_.
Page No 495:
Question 16:
If , then cot 2A = _________.
Answer:
Also,
Adding (1) and (2), we get
If , then cot 2A = ____0____.
Page No 495:
Question 17:
If ΔABC is an isosceles right triangle right angled at B, then
Answer:
It is given that ΔABC is an isosceles right triangle right angled at B.
.....(1)
In ΔABC,
(Angle sum property of a triangle)
= 1
If ΔABC is an isosceles right triangle right angled at B, then
Page No 495:
Question 18:
If α + β = 90° and , then tan α tan β = ________.
Answer:
Now,
If α + β = 90° and , then tan α tan β = ____1____.
Page No 495:
Question 19:
If in a triangle ABC, angles A and B are complementary, then the value of cot C is ________.
Answer:
It is given that angles A and B are complementary.
.....(1)
In âABC,
(Angle sum property of a triangle)
If in a triangle ABC, angles A and B are complementary, then the value of cot C is ____0____.
Page No 495:
Question 20:
The value of tan 25° tan 10° tan 80° tan 65° is __________.
Answer:
The value of tan 25° tan 10° tan 80° tan 65° is _____1_____.
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