Solution

Class 10 - Mathematics

TRIGONOMETRY REVISION ASSIGNMENT

1. ∵ △ PQR is a right angled triangle.
∴ PR2 + RQ2 = PQ2
⇒ PR2 = (13)2 - (12)2 = 25
⇒ PR = 5 cm
QR 12
i. (c) cos θ = =
PQ 13

1 13
ii. (c) sec θ = =
12
cos θ
PR 5
iii. (c) tan θ = =
12
...(i)
RQ
5 5

tan θ 60
∴
2
=
12

25
= 12

169
=
1+tan θ 169
1+
144 144

1 12
iv. (a) cotθ = = [Using (i)]
tan θ 5
PQ 13
cosecθ = =
PR 5

cot2θ cosec2θ
144 169
∴ - = − = -1
25 25

v. (b) sin2θ + cos2θ = 1 (Using identity)

–
2. i. (a) We have, AB = 9 m, BC = 3√3 m
In △ABC, we have
BC 3 √3 1
tan A = = =
AB 9 √3

⇒ tan A = tan30o ⇒ ∠A = 30o ...(i)

AB 9 –
ii. (c) Similarly, tan C = BC
= = √3
3 √3

−
−
⇒ tan C = tan60o ⇒ √C = 60o ...(ii)
sin 30o =
BC BC
iii. (d) Since, sin A = ⇒ [Using (i)]
AC AC

1
3 √3 –
⇒
2
=
AC
⇒ AC = 6√3 m

iv. (b) ∵ ∠A = 30o [From (i)]

cos 2A = cos(2 × 30o) = cos 60o =
1
∴
2

v. (b) ∴ ∠C = 60o [Using (ii)]

∘

= sin 30o =
C 60 1
∴ sin( 2
) = sin( 2
)
2

–
3. We have, KL = 4 cm, ML = 4√3 cm, KM = 8 cm
i. (a) tan M = KL
=
4
=
1

LM 4 √3 √3

⇒ tan M = tan30o ⇒ ∠ M - 30o

–
= tan 60o
ML 4 √3
ii. (c) tan K = = = √3
KL 4

⇒ ∠ K = 60o
1
iii. (b)
√3

iv. (c) tan2 M

2
2 ∘ (1 ) −1
tan 45 −1 0
v. (a) 2 ∘
= 2
=
2
=0
tan 45 +1 1 +1
–
4. We have, AB = BC = 6 √2 m and AC = 12 m.
i. (d) ∵ D is mid point of AC.
∴ AD = DC = 6 m

Now, AB2 = BD2 + AD2 (∵ △ ABD is a right triangle)

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 –
⇒ BD2 =(6√2)2 - 62 = 72 - 36 = 36
⇒ BD = 6 m ...(i)
BD 6 1
ii. (c) In △ABD, sin A = = = [Using (i)]
AB 6 √2 √2

⇒ sinA = sin45o ⇒ ∠A = 45o

BD 6
iii. (c) In △BDC, tan C = =
6
[Using (i)]
DC

⇒ tan C = 1 tan 45o ⇒ ∠C = 45o

, cosC = cos45o =
1 1
iv. (d) sinA =
√2 √2

1 1 2 –
∴ sinA + cosC = + = = √2
√2 √2 √2

v. (c) tanC = 1, tanA = tan45o = 1

⇒ tan2C + tan2A = 1 + 1 = 2
3
5. i. (a)
5

ii. (c) 60 m
iii. (b) 1
3
iv. (a) 4

v. (d) 1
–
6. i. (c) 4√3 m
ii. (a) 8 m
iii. (b) 16 m2
–
iv. (c) √3
1
v. (d) 2

AQ 1.2 3
7. i. (d) In △APQ, tanθ = = =
PQ 1.6 4

QB 3 15
ii. (d) In △PBQ, cot B = = = ...(i)
PQ 1.6 8

PQ 1.6
iii. (c) In △APQ, tan A = = =
4
...(ii)
AQ 1.2 3

iv. (d) We have, tan2A + 1 = sec2A

−−−−−−−
2
4
⇒ sec A = √( 3 ) + 1
−−−−− −−
16 25 5
=√ 9
+ 1 = √
9
=
3
−−−−−−− −
v. (a) Since, cosecB = √cot2 B + 1
−−−−−−−−
2
15
= √( ) + 1
8

17
= 8

1
8. i. (b) 2
–
ii. (a) 5√3 cm
iii. (a) ∠A
iv. (c) 60°
v. (b) BC
9. (b) Both A and R are true but R is not the correct explanation of A.
Explanation: For, θ = 45o, we have tan 45o = 1 and cot 45o = 1, so tan245 + cot245 = 1 + 1 = 2
10. (d) Assertion is wrong statement but reason is correct statement.
Explanation: cos A + cos2A = 1
cos A = 1 - cos2A = sin2A
sin2A + sin4A = cos A + cos2A = 1
sin2A + sin4A = 1
11. (d) A is false but R is true.
Explanation: sin θ and cosec θ are reciprocal of each other so sin θ ×  cosec θ = 1
sin θ × cosec θ ≠ cot θ

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12. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
P 4
Explanation: sinθ  =  H =
3

Here, perpendicular is greater than the hypotenuse which is not possible in any right triangle.
13. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
√3 –
Explanation: tanθ  =  2
× 2  = √3
14. (a) Both A and R are true and R is the correct explanation of A.
Explanation: If sin θ = cos θ then θ = 45o, so tan 45o = 1
15. (d) A is false but R is true.
1 Sin θ
Explanation: sin θ  cannot be equal to  . But tanθ  = 
Tan θ Cos θ

16. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
−−−−−−−−−
Explanation: Greatest side = √(3)2 + (4)2  = 5 units
17. (a) Both A and R are true and R is the correct explanation of A.
Explanation: As ∠ A + ∠ B + ∠ C = 180o and ∠ B = 90o
So, ∠ A + ∠ C = 90o
Hence, sin(A + C) = sin 90o = 1
18. (d) A is false but R is true.
Explanation: A is false but R is true.
19. (c) A is true but R is false.
Explanation: We know that, identity sin2 θ + cos2θ = 1 is true for every real value of θ so sin2θ = 1 – cos2θ
is always true.
20. (c) 1
Explanation: Given: x cosA = 1
1
⇒  x =  = sec A
cos A

And tan A = y
∴  x2 - y2 = sec2A - tan2A = 1
[∵  sec2θ  - tan2θ  = 1]
21. (d) 0
1
Explanation: Given: sinα = 
√2

⇒  sinα = sin45o
   = 45o
⇒ α

And tanβ = 1
⇒  tanβ = tan45o
⇒ β    = 45o
∴  cos(α + β) = cos(45o + 45o) = cos90o = 0
22. (a) 1
Explanation: Given,
sin θ + sin2 θ = 1
⇒ sin θ = 1 - sin2 θ
⇒ sin θ = cos2 θ
Now, cos2 θ + cos4 θ = sin θ + sin2 θ {∵ cos2 θ = sin θ }
⇒ cos2 θ + cos4 θ = 1
{∵ sin θ + sin2 θ = 1 (given)}
23. (b) 1
Explanation: If in right angled triangle PQR, right angled at Q, then P and R are acute angles.
Let ∠ P = θ, then ∠ R

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 = 90o - θ
Now, sin(P + R) = sin(θ  + 90o - θ ) = sin90o = 1

24. (c) tan2θ  + sin2θ

Explanation: Given: (secθ  + cosθ )(secθ  - cosθ )
= (sec2θ  - cos2θ )
= (1 + tan2θ  - 1 + sin2θ )
= (tan2θ  + sin2θ )
–
25. (a) √2 sin θ
–
Explanation: Given: sinθ  + cosθ  = √2 cosθ
Squaring both sides, we get
⇒  sin2θ  + cos2θ  + 2 sinθ  cosθ  = 2 cos2θ
⇒  cos2θ  - 2sinθ  cosθ  = sin2θ
⇒  cos2θ  - 2sinθ  cosθ  + sin2θ  = 2sin2θ
⇒  (cosθ  - sinθ )2 = 2sin2θ
–
⇒  cosθ  - sinθ  = √2 sinθ
2
m −1
26. (a) 2
n −1

Explanation: Given: tanA = n tanB

1 n
⇒   =
tan B tan A
n
 
⇒ cot B =
tan A

And sin A = m sin B

1 m
⇒   sin B =
sin A
n
⇒  cosec B = 
sin A

Now, cosec2B - cot2B = 1

2 2
m n
⇒   2
− 2
 = 1
sin A tan A
2 2 2
m n cos A
⇒ 2
− 2
 = 1
sin A sin A

⇒  m2 - n2cos2A = sin2A

⇒  m2 - n2cos2A = 1 - cos2A
⇒  m2 - 1 = (n2 - 1)cos2A
2

 cos2A = 
m −1
⇒ 2
n −1

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