Page No 546:
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 546:
Question 2:
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 =
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 546:
Question 3:
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 =
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 546:
Question 4:
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 =
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 546:
Question 5:
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 =
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 = 10k2k2
⇒ BC2 = 9k2
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
tan = =
cos =
∴ , cot = and sec =
Page No 546:
Question 6:
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 = 10k2k2
⇒ BC2 = 9k2
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
tan = =
cos =
∴ , cot = and sec =
Answer:
We have ,
As,
Also,
Now,
Also,
And,
Page No 546:
Question 7:
We have ,
As,
Also,
Now,
Also,
And,
Answer:
Page No 546:
Question 8:
Answer:
Page No 546:
Question 9:
Answer:
Page No 546:
Question 10:
Answer:
Page No 547:
Question 11:
Answer:
It is given that tan .
LHS =
Dividing the numerator and denominator by cos , we get:
(âµ tan )
Now, substituting the value of tan in the above expression, we get:
i.e., LHS = RHS
Hence proved.
Page No 547:
Question 12:
It is given that tan .
LHS =
Dividing the numerator and denominator by cos , we get:
(âµ tan )
Now, substituting the value of tan in the above expression, we get:
i.e., LHS = RHS
Hence proved.
Answer:
Page No 547:
Question 13:
Answer:
Page No 547:
Question 14:
Answer:
Page No 547:
Question 15:
Answer:
It is given that cot .
LHS =
Dividing the above expression by sin , we get:
[âµ cot ]
Now, substituting the values of cot in the above expression, we get:
i.e., LHS = RHS
Hence proved.
Page No 547:
Question 16:
It is given that cot .
LHS =
Dividing the above expression by sin , we get:
[âµ cot ]
Now, substituting the values of cot in the above expression, we get:
i.e., LHS = RHS
Hence proved.
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 547:
Question 17:
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.
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 547:
Question 18:
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.
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 547:
Question 19:
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.
Answer:
Page No 547:
Question 20:
Answer:
(i)
â
(ii)
Page No 547:
Question 21:
(i)
â
(ii)
Answer:
Page No 547:
Question 22:
Answer:
Page No 547:
Question 23:
Answer:
Page No 547:
Question 24:
Answer:
Page No 548:
Question 25:
Answer:
Page No 548:
Question 26:
Answer:
In
,
Using Pythagoras theorem, we get
Now,
Page No 548:
Question 27:
In
,
Using Pythagoras theorem, we get
Now,
Answer:
Page No 548:
Question 28:
Answer:
Page No 548:
Question 29:
Answer:
In
ABC,
C = 90
sinA =
and
sinB =
As, sinA = sinB
=
BC = AC
So,
A =
B (Angles opposite to equal sides are equal)
Page No 548:
Question 30:
In
ABC,
C = 90
sinA =
and
sinB =
As, sinA = sinB
=
BC = AC
So,
A =
B (Angles opposite to equal sides are equal)
Answer:
In
,
Page No 555:
Question 1:
In
,
Answer:
Hence, the correct option is (c).
Page No 555:
Question 2:
Hence, the correct option is (c).
Answer:
Hence, the correct option is (a).
Page No 555:
Question 3:
Hence, the correct option is (a).
Answer:
Hence, the correct option is (d).
Page No 555:
Question 4:
Hence, the correct option is (d).
Answer:
Hence, the correct option is (b).
Page No 555:
Question 5:
Hence, the correct option is (b).
Answer:
Hence, the correct option is (c).
Page No 555:
Question 6:
Hence, the correct option is (c).
Answer:
Hence, the correct option is (a).
Page No 556:
Question 7:
Hence, the correct option is (a).
Answer:
Hence, the correct option is (c).
Page No 556:
Question 8:
Hence, the correct option is (c).
Answer:
Page No 556:
Question 9:
Answer:
Page No 556:
Question 10:
Answer:
Page No 556:
Question 11:
Answer:
Page No 556:
Question 12:
Answer:
Page No 556:
Question 13:
Answer:
Page No 556:
Question 14:
Answer:
Page No 556:
Question 15:
Answer:
Page No 556:
Question 16:
Answer:
Hence, the correct option is (a).
View NCERT Solutions for all chapters of Class 10