Rs Aggarwal 2015 Solutions for Class 10 Math Chapter 5 Trigonometric Ratios are provided here with simple step-by-step explanations. These solutions for Trigonometric Ratios are extremely popular among Class 10 students for Math Trigonometric Ratios Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2015 Book of Class 10 Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2015 Solutions. All Rs Aggarwal 2015 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.
Page No 273:
Question 1:
If sin , find the value of all T-ratios of θ.
Answer:
Let us first draw a right ABC, right angled at B and .
Now, we know that sin = = = .
So, if AB = , then AC = 2k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2 AB2 = (2k)2 ()2
⇒ BC2 = 4k2 3k2 = k2
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:
cos = =
tan =
∴ cot = , cosec = and sec =
Page No 273:
Question 2:
If cos find the values of all T-ratios of θ.
Answer:
Let us first draw a right ABC, right angled at B and .
Now, we know that cos = = = .
So, if BC = 7k, then AC = 25k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 BC2 = (25k)2 (7k)2.
⇒ AB2 = 625k2 49k2 = 576k2
⇒ AB = 24k
Now, finding the other trigonometric ratios using their definitions, we get:
sin = =
tan =
∴ cot = , cosec = and sec =
Page No 273:
Question 3:
If tan find the values of all T-ratios of θ.
Answer:
Let us first draw a right ABC, right angled at B and .
Now, we know that tan = = = .
So, if BC = 8k, then AB = 15k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (15k)2 + (8k)2
⇒ AC2 = 225k2 + 64k2 = 289k2
⇒ AC = 17k
Now, finding the other T-ratios using their definitions, we get:
sin = =
cos =
∴ cot = , cosec = and sec =
Page No 273:
Question 4:
If cot θ = 2, find the value of all T-ratios of θ.
Answer:
Let us first draw a right ABC, right angled at B and .
Now, we know that cot = = = 2.
So, if BC = 2k, then AB = k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (2k)2 + (k)2
⇒ AC2 = 4k2 + k2 = 5k2
⇒ AC = k
Now, finding the other T-ratios using their definitions, we get:
sin = =
cos =
∴ tan = , cosec = and sec =
Page No 273:
Question 5:
If cosec θ = , the find the values of all T-ratios of θ.
Answer:
Let us first draw a right ABC, right angled at B and .
Now, we know that cosec = = = .
So, if AC = ()k, then AB = k, where k is a positive number.
Now, by using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2 AB2 = 10k2 k2
⇒ BC2 = 9k2
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
tan = =
cos =
∴ , cot = and sec =
Page No 273:
Question 6:
If cot θ = , show then evaluate
Answer:
Let us consider a right ABC, right angled at B.
Now, we know that cot = = = .
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (8k)2 + (15k)2
⇒ AC2 = 64k2 + 225k2 = 289k2
⇒ AC = 17k
Finding out the value of sin and cos using their definitions, we have:
sin =
cos =
Substituting these values in the given expression, we get:
Page No 273:
Question 7:
If tan θ = , show that (sin θ + cos θ) = .
Answer:
Let us consider a right ABC, right angled at B and .
Now, we know that tan = = .
So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (4k)2 + (3k)2
⇒ AC2 = 16k2 + 9k2 = 25k2
⇒ AC = 5k
Finding out the values of sin and cos using their definitions, we have:
sin =
cos =
Substituting these values in the given expression, we get:
(sin + cos ) =
i.e., LHS = RHS
Hence proved.
Page No 273:
Question 8:
If tan θ = , show that
Answer:
Let us consider a right ABC, right angled at B and .
Now it is given that tan = = .
So, if AB = k, then BC = k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AC2 = (k)2 + (k)2
⇒ AC2 = k2 + 7k2
⇒ AC = 2k
Now, finding out the values of the other trigonometric ratios, we have:
sin =
cos =
∴ cosec = and sec =
Substituting the values of cosec and sec in the given expression, we get:
i.e., LHS = RHS
Hence proved.
Page No 273:
Question 9:
If cosec θ = 2, show that
Answer:
Let us consider a right ABC, right angled at B and .
Now, it is given that cosec = 2.
Also, sin =
So, if AB = k, then AC = 2k, where k is a positive number.
Using Pythagoras theorem, we have:
⇒ AC2 = AB2 + BC2
⇒ BC2 = AC2 AB2
⇒ BC2 = (2k)2 (k)2
⇒ BC2 = 3k2
⇒ BC = k
Finding out the other T-ratios using their definitions, we get:
cos =
tan =
cot =
Substituting these values in the given expression, we get:
i.e., LHS = RHS
Hence proved.
Page No 273:
Question 10:
If sec θ = , verify that .
Answer:
Let us consider a right ABC, right angled at B and .
Now, it is given that sec = .
Also, cos =
So, if BC = 4k, then AC = 5k, where k is a positive number.
Using Pythagoras theorem, we have:
⇒ AC2 = AB2 + BC2
⇒ AB2 = AC2 BC2
⇒ AB2 = (5k)2 (4k)2 = 25k2 16k2
⇒ AB2 = 9k2
⇒ AB = 3k
Finding out the other T-ratios using their definitions, we get:
sin =
tan =
Substituting the values in LHS and RHS, we get:
i.e., LHS = RHS
Hence proved.
Page No 273:
Question 11:
If cos θ = 0.6, show that (5 sin θ − 3 tan θ) = 0.
Answer:
Let us consider a right ABC right angled at B.
Now, we know that cos = 0.6 = =
So, if BC = 3k, then AC = 5k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 BC2
⇒ AB2 = (5k)2 (3k)2 = 25k2 9k2
⇒ AB2 = 16k2
⇒ AB = 4k
Finding out the other T-ratios using their definitions, we get:
sin =
tan =
Substituting the values in the given expression, we get:
5 sin 3 tan
i.e., LHS = RHS
Hence proved.
Page No 273:
Question 12:
If 3 tan θ = 4, show that
Answer:
Let us consider a right ABC right angled at B and .
We know that tan = =
So, if BC = 3k, then AB = 4k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AC2 = 16k2 + 9k2
⇒ AC2 = 25k2
⇒ AC = 5k
Now, we have:
sin =
cos =
Substituting these values in the given expression, we get:
i.e., LHS = RHS
Hence proved.
Page No 273:
Question 13:
If tan θ = , show that
Answer:
It is given that tan .
LHS =
Dividing the numerator and denominator by cos , we get:
(∵ tan )
Now, substituting the value of tan in the above expression, we get:
i.e., LHS = RHS
Hence proved.
Page No 273:
Question 14:
If 3 cot θ = 2, show that
Answer:
It is given that cot .
LHS =
Dividing the above expression by sin , we get:
[∵ cot ]
Now, substituting the values of cot in the above expression, we get:
i.e., LHS = RHS
Hence proved.
Page No 273:
Question 15:
If sec θ = , verify that .
Answer:
It is given that sec = .
Let us consider a right ABC right angled at B and .
We know that cos =
So, if BC = 8k, then AC = 17k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 BC2 = (17k)2 (8k)2
⇒ AB2 = 289k2 64k2 = 225k2
⇒ AB = 15k.
Now, tan = and sin =
The given expression is .
Substituting the values in the above expression, we get:
∴ LHS = RHS
Hence proved.
Page No 273:
Question 16:
If tan θ = , show that
Answer:
Let us consider a right ABC right angled at B and .
Now, we know that tan = =
So, if AB = 20k, then BC = 21k, where k is a positive number.
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2= (20k)2 + (21k)2
⇒ AC2 = 841k2
⇒ AC = 29k
Now, sin = and cos =
Substituting these values in the given expression, we get:
∴ LHS = RHS
Hence proved.
Page No 273:
Question 17:
In a ∆ABC, right-angled at B, it is given that AB = 12 cm and BC = 5 cm.
Find the value of
(i) cos A
(ii) cosec A
(iii) cos C
(iv) coses C.
Answer:
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2 = 122 + 52 = 144 + 25
⇒ AC2 = 169
⇒ AC = 13 cm
Now, for T-Ratios of ∠A, base = AB and perpendicular = BC
(i) cos A =
(ii) cosec A =
Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) cos C =
(iv) cosec C =
Page No 274:
Question 18:
In a ∆ABC, it is given that ∠B = 90°, AB = 24 cm and BC = 7 cm.
Figure
Find the value of
(i) sin A
(ii) cos A
(iii) sin C
(iv) cos C
Answer:
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2= (24)2 + (7)2
⇒ AC2 = 576 + 49 = 625
⇒ AC = 25 cm
Now, for T-Ratios of ∠A, base = AB and perpendicular = BC
(i) sin A =
(ii) cos A =
Similarly, for T-Ratios of ∠C, base = BC and perpendicular = AB
(iii) sin C =
(iv) cos C =
Page No 274:
Question 19:
Given a ∆ABC, in which ∠C = 90°, ∠ABC = θ°, BC = 21 units, AB = 29 units.
Show that (cos2θ − sin2θ) =
Answer:
Using Pythagoras theorem, we get:
AB2 = AC2 + BC2
⇒ AC2 = AB2 BC2
⇒ AC2 = (29)2 (21)2
⇒ AC2 = 841 441
⇒ AC2 = 400
⇒ AC = = 20 units
Now, sin and cos =
cos2 sin2 =
Hence Proved.
Page No 276:
Question 1:
If cosθ = , then tanθ = ?
(a)
(b)
(c)
(d)
Answer:
(a)
Since cosθ = but cosθ =
So, =
Thus, AB = 4k and AC = 5k
Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2 AB2
⇒ BC2 = (5k)2 (4k)2
⇒ BC2 = 9k2
⇒ BC = 3k
∴ tanθ = =
Page No 276:
Question 2:
If sinθ = , then cotθ = ?
(a)
(b)
(c)
(d) 1
Answer:
(b)
Given: sinθ = , but sinθ =
So, =
Thus, BC = k and AC = 2k
Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
AB2 = AC2 BC2
AB2 = (2k)2 (k)2
AB2 = 3k2
AB = k
So, tanθ = = =
∴ cotθ = =
Page No 276:
Question 3:
If tan θ = , then cosec θ = ?
(a)
(b)
(c)
(d)
Answer:
(d)
Given: tan = , but tan =
So, let us first draw a right ABC right angled at B and .
So, =
Thus, BC = 3k and AB = 4k
Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AC2 = (4k)2 + (3k)2
⇒ AC2 = 25k2
⇒ AC = 5k
∴ cosec = = =
Page No 276:
Question 4:
If cosec θ = , then sec θ = ?
(a)
(b)
(c)
(d)
Answer:
(b)
Let us first draw a right ABC right angled at B and .
Given: cosec = , but sin = =
Also, sin = =
So, =
Thus, BC = k and AC = k
Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 BC2
⇒ AB2 = ( k)2 (k)2
⇒ AB2 = 9k2
⇒ AB = 3k
∴ sec = =
Page No 276:
Question 5:
If tan θ = , then cosec θ = ?
(a)
(b)
(c)
(d)
Answer:
(a)
Let us first draw a right ABC right angled at B and .
Given: tan = , but tan =
So, =
Thus, BC = 8k and AB = 15k
Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AC2 = (15k)2 + (8k)2
⇒ AC2 = 289k2
⇒ AC = 17k
∴ cosec =
Page No 276:
Question 6:
If sin θ , then cos θ = ?
(a)
(b)
(c)
(d)
Answer:
(b)
Let us first draw a right ABC right angled at B and .
Given: sin θ = , but sin θ =
So, =
Thus, BC = ak and AC = bk
Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 BC2
⇒ AB2 = (bk)2 (ak)2
⇒ AB2 = k2
⇒ AB = ()k
∴ cos θ = = =
Page No 276:
Question 7:
If tan θ = , then sec θ = ?
(a)
(b)
(c)
(d) 2
Answer:
(d) 2
Let us first draw a right ABC right angled at B and .
Given: tan θ =
But tan θ =
So, =
Thus, BC = k and AB = k
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2 = ( k)2 + (k)2
⇒ AC2= 4k2
⇒ AC = 2k
∴ sec θ =
Page No 276:
Question 8:
If sec θ = , then sin θ = ?
(a)
(b)
(c)
(d) None of these
Answer:
(c)
Let us first draw a right ABC right angled at B and .
Given: sec θ =
But cos θ = = =
Thus, AC = 25k and AB = 7k
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ BC2 = AC2 AB2
⇒ BC2 = (25k)2 (7k)2
⇒ BC2= 576k2
⇒ BC = 24k
∴ sin θ =
Page No 276:
Question 9:
If sin θ = , then (cosecθ + cotθ) = ?
(a)
(b)
(c)
(d)
Answer:
(d)
Let us first draw a right ABC right angled at B and .
sin θ = =
So, BC = k and AC = 2k
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AB2 = AC2 BC2
⇒ AB2 = (2k)2 (k)2
⇒ AB2 = k2
⇒ AB = k
So, cot θ = =
cosec θ =
∴ (cosec θ + cot θ) = =
Page No 277:
Question 10:
If tan θ = , then (sinθ + cosθ) = ?
(a)
(b)
(c)
(d)
Answer:
(c)
Let us first draw a right ABC right angled at B and .
tan θ =
So, AB = 3k and BC = 4k
Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2 = (3k)2 + (4k)2
⇒ AC2 = 25k2
⇒ AC= 5k
Thus, sin θ = =
and cos θ =
∴ (sin θ + cos θ) = ( + ) =
Page No 277:
Question 11:
If 3x = cosec θ and = cot θ, than
(a)
(b)
(c)
(d)
Answer:
(c)
Given: 3x = cosec θ and = cot θ
Also, we can deduce that x = and .
So, substituting the values of x and in the given expression, we get:
3 = 3
= 3
=
= [By using the identity: ]
Page No 277:
Question 12:
If 2x = sec A and = tan A, then =?
(a)
(b)
(c)
(d)
Answer:
(a)
Given: 2x = sec A and = tan A
Also, we can deduce that x = and .
So, substituting the values of x and in the given expression, we get:
2 = 2
= 2
=
= [By using the identity: ]
Page No 277:
Question 13:
If (tan θ + cot θ) = 5, then (tan2 θ + cot2 θ) = ?
(a) 27
(b) 25
(c) 24
(d) 23
Answer:
(d) 23
We have (tan θ +cot θ) = 5
Squaring both sides, we get:
(tan θ +cot θ)2 = 52
⇒ tan2 θ + cot2 θ + 2 tan θ cot θ = 25
⇒ tan2 θ + cot2 θ + 2 = 25 [∵ tan θ = ]
⇒ tan2 θ + cot2 θ = 25 − 2 = 23
Page No 277:
Question 14:
If (cos θ + sec θ) = , then (cos2 θ + sec2 θ) = ?
(a)
(b)
(c)
(d)
Answer:
(b)
We have (cos θ +sec θ) =
Squaring both sides, we get:
(cos θ + sec θ)2 = ()2
cos2 θ + sec2 θ + 2 cos θ sec θ =
cos2 θ + sec2 θ + 2 = [∵ sec θ = ]
cos2 θ + sec2 θ = − 2 =
Page No 277:
Question 15:
If tan θ = , then = ?
(a)
(b)
(c)
(d)
Answer:
(d)
=
Page No 277:
Question 16:
If 7 tan θ = 4, then = ?
(a)
(b)
(c)
(d)
Answer:
(a)
7 tan θ = 4
Now, dividing the numerator and denominator of the given expression by cos θ, we get:
=
= [∵ 7 tan θ = 4]
=
Page No 277:
Question 17:
If 3 cot θ = 4, then = ?
(a)
(b) 3
(c)
(d) 9
Answer:
(d) 9
We have .
Dividing the numerator and denominator of the given expression by sin θ, we get:
=
= = 9 [∵ 3 cot θ = 4]
Page No 277:
Question 18:
If θ = , then = ?
(a)
(b)
(c)
(d)
Answer:
(b)
We have tan θ =
Now, dividing the numerator and denominator of the given expression by cos θ, we get:
Page No 278:
Question 19:
If cot θ = , then = ?
(a)
(b)
(c)
(d)
Answer:
(b)
We have:
Page No 278:
Question 20:
In the given ∆ABC, it is given that ∠B = 90°, AB = 24 cm and BC = 7 cm. Then, cos A = ?
Figure
(a)
(b)
(c)
(d) None of thse
Answer:
(c)
Here, AB = 24 cm and BC = 7 cm
Now, using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2 = (24)2 + (7)2
⇒ AC2 = 625
⇒ AC = 25 cm
∴ cos A =
View NCERT Solutions for all chapters of Class 10