KSEEB SSLC Solutions for Class 10 Maths Chapter 12 Trigonometry

In this chapter, we provide KSEEB Solutions for Class 10 Maths Chapter 12 Trigonometry for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB Solutions for Class 10 Maths Chapter 12 Trigonometry pdf, free KSEEB solutions for Class 10 Maths Chapter 12 Trigonometry book pdf download. Now you will get step by step solution to each question.

KSEEB SSLC Solutions for Class 10 Maths – Trigonometry (English Medium)

KSEEB SSLC Solutions for Class 10 Maths Chapter 12 Exercise 12.1

Question I:

Find sinθ and cosθ for the following:

Solution :

Question II:

Solution :

Question III(1):

Solution :

Question III(2):

Solution :

Question III(3):

Solution :

Question III(4):

Solution :

Question III(5):

If 3 tanθ = 1, find sinθ, cosθ and cotθ.

Solution :

Question III(6):

If sec x = 2, then find sin x, tan x and cot x and cot x + cosec x

Solution :

Question III(7):

If 4sinA – 3cosA = 0, find sin A, cos A and sec A, cosec A

Solution :

Question III(8):

Solution :

Question III(9):

Solution :

Question III(10):

Solution :

Exercise 12.2:

Question I:

Answer the following questions:

1. What trigonometric ratios of angles from 0° to 90° are equal to 0?
2. What trigonometric ratios of angles from 0° to 90° are equal to 1?
3. What trigonometric ratios of angles from 0° to 90° are equal to 0.5?
4. What trigonometric ratios of angles from 0° to 90° are not defined?
5. What trigonometric ratios of angles from 0° to 90° are equal?

Solution :

1. The trigonometric ratios of angles from 0° to 90° are equal to 0 are sin0°, cos90°, tan0°, cot90°.
2. The trigonometric ratios of angles from 0° to 90° which are equal to 1 are sin90°, cos0°, tan45°, cosec90°, sec0°, cot45°.
3. The trigonometric ratios of angles from 0° to 90° which are equal to 0.5 are sin30° and cos60°.
4. The trigonometric ratios of angles from 0° to 90° which are not defined are tan90°, cosec0°, sec90°, cot0°.
5. The trigonometric ratios of angles from 0° to 90° which are equal are sin45° and cos45°, sec45° and cosec45°, tan45° and cos45°, sin90° and cosec90°, cos0° and sec0°.

Question II:

Solution :

Question III(i):

Find the value of the following:
sin30° × cos60° – tan²45°

Solution :

Question III(ii):

Find the value of the following:
sin60° × cos30° + cos60° × sin30°

Solution :

Question III(iii):

Find the value of the following:
cos60° × cos30° – sin60° × sin30°

Solution :

Question III(iv):

Find the value of the following:
2sin²30° – 3cos²30° + tan60° + 3sin²90°.

Solution :

Question III(v):

Find the value of the following:
4sin²60° + 3tan²30° – 8sin45° × cos45°

Solution :

Question III(vi):

Solution :

Question III(vii):

Solution :

Question III(viii):

Solution :

Question III(ix):

Solution :

Question III(x):

Solution :

Question IV(i):

Prove the following equality
sin30° × cos60° + cos30° × sin60° = sin90°

Solution :

Question IV(ii):

Prove the following equality
2cos²30° – 1 = 1 – 2sin²30°=cos60°

Solution :

Question IV(iii):

Prove the following equality
If θ = 30° prove that 4cos²θ – 3cos0° = cos3θ

Solution :

Question IV(iv):

Solution :

Question IV(v):

Prove the following equality If B = 15°, prove that 4 sin2B × cos4B × sin6B = 1

Solution :

Question IV(vi):

Solution :

KSEEB SSLC Solutions for Class 10 Maths Chapter 12 Exercise 12.3

Question I(1):

Show that
(1 – sin²θ) sec²θ = 1

Solution :

Question I(2):

Show that
(1 + tan²θ) cos²θ = 1

Solution :

Question I(3):

Show that
(1 + tan²θ)(1 – sinθ)(1 + sinθ)=1

Solution :

Question I(4):

Solution :

Question I(5):

Solution :

Question I(6):

Show that
(1 + cosA)(1 – cosA)(1 + cot²A) = 1

Solution :

Question I(7):

Solution :

Question I(8):

Solution :

Question I(9):

Show that
(sin θ +cos θ)² = 1 + 2sinθcosθ

Solution :

L.H.S. = (sin θ + cos θ)²
 = sin²θ + 2sinθcosθ + cos²θ
= 1 + 2sinθcosθ
= R.H.S.
∴(sin θ +cos θ)² = 1 + 2sinθcosθ

Question I(10):

Solution :

Question I(11):

Show that
Sin A cos A tan A + cos A sin A cot A = 1

Solution :

Question I(12):

Solution :

Question I(13):

Solution :

Question I(14):

Show that
tan²A – sin²A = tan²Asin²A

Solution :

Question I(15):

Show that
cos²A – sin²A = 2cos²A – 1

Solution :

Question I(16):

Show that
If x = r sin A cos B, y = r sin A sin B, z = r cos A, then x² + y² + z² = r²

Solution :

Exercise 12.4:

Question 1(i):

Solution :

Question 1(ii):

Solution :

Question 1(iii):

Evaluate:
cos48°- sin42°

Solution :

cos48° – sin42°
= sin(90° – 48°) – sin42°
= sin42° – sin42°
= 0

Question 1(iv):

Evaluate:
cosec31° – sec59°

Solution :

cosec31° – sec59°
= sec(90° – 31°) – sec59°
= sec59° – sec59°
= 0

Question 1(v):

Evaluate:
cot34° – tan56°

Solution :

cot34° – tan56°
= tan(90° – 34°) – tan56°= tan56° – tan56°
= 0

Question 1(vi):

Solution :

Question 1(vii):

Evaluate:
sec70° sin20° – cos70° cosec20°

Solution :

sec70° sin20° – cos70° cosec20°
= sec(90° – 20°)sin(90° – 70°) – cos70°cosec20°
= cosec20° cos70° – cos70°cosec 20°
= 0

Question 1(viii):

Evaluate:
cos²13° – sin²77°

Solution :

cos²13° – sin²77°
= cos²(90° – 77°) – sin²77
= sin² 77 – sin² 77
= 0

Question 2(i):

Prove that
sin35° sin55° – cos35° cos55°= 0

Solution :

L.H.S. = sin35° sin55° – cos35° cos55°= sin(90° – 55°)sin55° – cos(90° – 55°)cos55°
= cos55° sin55°- sin55° cos55°
= 0
= R.H.S.
∴ sin35° sin55° – cos35° cos55° = 0

Question 2(ii):

Prove that
tan10° tan15° tan75° tan80°= 1

Solution :

Question 2(iii):

Prove that
cos38° cos52° – sin38° sin52°= 0

Solution :

L.H.S. = cos38° cos52° – sin38° sin52°
= cos(90° – 52°) cos(90° – 38°) – sin38° sin52°
= sin52° sin38° – sin38° sin52°
= 0
= R.H.S.
∴ cos38° cos52° – sin38° sin52° = 0

Question 3:

If sin 5θ = cos 4θ, where 5θ and 4θ are acute angles, find the value of θ.

Solution :

sin5θ = cos4θ
∴ cos(90° – 5θ) = cos4θ
∴ 90° – 5θ = 4θ
∴ 90° = 4θ + 5θ
∴ 9θ = 90°

Question 4:

It tan2A = cot(A – 18°), where 2A is an acute angle, find the value of A.

Solution :

Question 5:

If sec4A = cosec(A – 20°), where 4A is an acute angle, find the value of A.

Solution :

Exercise 12.5:

Question I(1):

Find the value of ‘x’.

Solution :

Question I(2):

Find the value of ‘x’.

Solution :

Question I(3):

Find the value of ‘x’.

Solution :

Question I(4):

Find the value of ‘x’.

Solution :

Question I(5):

Find the value of ‘x’.

Solution :

Question II(1):

A tall building casts a shadow of 300 m long when the sun’s altitude (elevation) is 30°. Find the height of the tower.

Solution :

Question II(2):

Solution :

Question II(3):

A tree is broken over by the wind forms a right angled triangle with the ground. If the broken part makes an angle of 60° with the ground and the top of the tree is now 20 m from its base, how tall was the tree?

Solution :

Question II(4):

The angle of elevation of the top of a flag post from a point on a horizontal ground is found to be 30°.  On walking 6 m towards the post, the elevation increased by 15°. Find the height of the flag post.

Solution :

Question II(5):

Two observers are 1 km apart in the same vertical plane as a balloon, but on opposite sides of it, found the angles of elevation to be 60° and 45°. How high is the balloon above the ground.

Solution :

Question II(6):

The angles of elevation of the top of a cliff as seen from the top and bottom of a building are 45° and 60° respectively. If the height of the building is 24 m, find the height of the cliff.

Solution :

Question II(7):

From the top of a building 16 metre high, the angular elevation of the top of a hill is 60° and the angular depression of the foot of the hill is 30°. Find the height of the hill.

Solution :

Question II(8):

Solution :

Question II(9):

From a point 50 m above the ground the angle of elevation of a cloud is 30° and the angle of depression of its reflection is 60°. Find the height of the cloud above the ground.

Solution :

Question II(10):

From a light house the angles of depression of two ships on opposite sides of the light house were observed to be 45° and 60°. If the height of the light house is 120 m and the line joining the two ships passes through the foot of the light house, find the distance between to two ships.

Solution :

All Chapter KSEEB Solutions For Class 10 Maths

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All Subject KSEEB Solutions For Class 10

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