NCERT Grade 10-Introduction to Trigonometry-Answers

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 1.    Let us draw a right angled triangle ABC, right angled at B.

Using Pythagoras theorem.

AC2 = AB2 + BC2

= (24)2+ (7)2 = 576 + 49 = 625

⇒ AC = 25 cm

2.   Using Pythagoras theorem,

PR2 = PQ2 + QR2                ⇒           (13)2 = (12)2+ QR2

⇒ QR2 =169 – 144 = 25       ⇒          QR = 5 cm

3.   Given: A triangle ABC in which ∠B = 900

Let BC = 3k and AC = 4k

Then, Using Pythagoras theorem,

4.   Given: A triangle ABC in which ∠B = 900

Let AB = 8k and BC = 15k

Then using Pythagoras theorem,

5.   Consider a triangle ABC in which ∠ A = 9 and ∠ B = 900

Let AB = 12k and BC = 5k

Then, using Pythagoras theorem,

6.   In right triangle ABC,

But      cos A = cos B         [Given]

[Angles opposite to equal sides are equal]

7.    Consider a triangle ABC in which ∠ A = 9 and ∠ B = 900 Let AB = 7k and BC = 8k

Then, using Pythagoras theorem,

8.    Consider a triangle ABC in which ∠ B = 900.

Let AB = 4k and BC = 3k.

Then, using Pythagoras theorem,

9.   Consider a triangle ABC in which ∠ B = 900

Let BC = and AB = √3k

Then, using Pythagoras theorem,

For ∠ C, Base = BC, Perpendicular = AB and Hypotenuse = AC

10.   In Δ PQR, right angled at Q.

PR + QR = 25 cm and PQ = 5 cm

Let           QR = x cm and PR = (25 –x) cm

Using Pythagoras theorem,

RP2 = RQ2 + QP2                        ⇒     (25 – x)2 = (x)2 + (5)2

⇒ 625 -50x + x2 = x2 + 25        ⇒      -50x = -600          ⇒        x =12

∴ RQ = 12 cm and RP = 25 – 12 = 13 cm

11.  (i) False because sides of a right triangle may have any length, so tan A may have any value.

(ii) True as sec A is always greater than 1.

(iii) False as cos A is the abbreviation of cosine A.

(iv) False as cot A is not the product of ‘cot’ and A. ‘cot’ is separated from A has no meaning.

(v) False as sin θ cannot be > 1.

12.   Solution:

13.   Solution:

(iii)    (A) Since A = 0, then sin 2 A = sin 00 = 0 and 2 sin A = 2 sin 00 = 2 × 0 = 0

∴ sin 2 A = sin A when A = 0

14.   

On adding eq. (i) and (ii), we get,

2A = 900          ⇒        A = 450

On Subtracting eq. (i) and eq. (ii), we get

2B = 300          ⇒          B = 150

15.   Solutions:

(i) False, because sin (A + B) = sin (600 + 300) = sin 900 = 1

And            sin A + sin B = sin 600 + sin 300

∴                   sin (A + B) ≠ sin A + sin B

(ii) True, because

It is clear, the value of cos θ increases as θ increases.

(iii) False, because

It is clear, the value of cos θ increases as θ increases.

(iv) False as it is only true for 

(v) True, because  undefined.

16.   Solution:

17.  (i)     tan 480 tan 230 tan 420 tan 670

 = tan (900 – 420) tan (900 – 670) tan 420 tan 670 = cot 420 cot 670 tan 420 tan 670

 tan 420. tan 670 = 1 = R.H.S.

(ii)   cos 380 cos 520 – sin 38sin 520

= cos (900 – 52). cos (900 – 380) – sin 380. sin 520

= sin 520 sin 380 – sin 380 sin 520 = 0 = R.H.S.

18.   Given: tan 2A = cot (A -18°)

⇒ cot (900 – 2A) = cot (A -180)     ⇒    900 – 2A = A -180

⇒ -2A – A = -180 – 900                    ⇒    -3A = -1080

⇒ A= 360

19.   Given: tan A = cot B

⇒ cot (900 – A) = cot B             ⇒        900 – A = B

⇒ A – FB = 900

20.   Given: sec 4A = cos ec (A- 20°)

⇒ cos ec(900 – 4A) = cos ec(A – 200)        ⇒       900 – 4A = A -200

⇒ -4A – A = -200 – 900          ⇒        -5 A = -1100

⇒ A = 220

21.  Given: A, B and C are interior angles of a Δ ABC.

22.   sin 67+ cos 750 = sin(90— 231 + cos (90– 150) = cos 23+  sin 150

23.  For sin A,

By using identity cos ec2 A – cot2 A = 1      ⇒       cos ec2 A = l + cot2

For sec A,

By using identity sec2 A – tan2 A = 1     ⇒       sec2 A = 1 + tan2 A

For tan A,

24.   For sin A,

By using identity, sin2 A + cos2 A = 1              sin2 A = 1 – cos2 A

For cos A,

For tan A,

By using identity sec2 A – tan2 A = 1    ⇒     tan2 A= sec2 A – 1

For cos ecA,

For cot A,

25.   

(ii) sin 250 cos 650 + cos 250 sin 650 = sin 25.cos (900 — 250) + cos 250. sin (900 — 250)

= sin 250. sin 250 + cos 250. cos 250         [∵ sin (900θ) = cos θ, cos (900θ) = sin θ]

= sin2 250 + cos2 250 = 1           [∵ sin2 θ + cos2 θ = 1]

26. 

27.   Proof:

(i) L.H.S.    (cos ecθ — cot θ)2 = cos ec2θ + cot2 θ — 2 cos ecθ cot θ

(ii)

(iii)

 

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(x)

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