Unit Test (Solutions): Introduction to Trigonometry | Mathematics (Maths) Class 10 PDF Download

Time: 1 hour

M.M. 30

Attempt all questions.

• Question numbers 1 to 5 carry 1 mark each.
• Question numbers 6 to 8 carry 2 marks each.
• Question numbers 9 to 11 carry 3 marks each.
• Question number 12 & 13 carry 5 marks each.

Q1: When cos A = 4/5, the value for tan A is (1 Mark)
(a) 3/5
(b) 3/4
(c) 5/3
(d) 4/3

Ans: (b) 3/4

Q2: Which of the following is the the simplest value of cos² θsin θ + sin θ (1 Mark)

(a) cosec θ
(b) sec θ
(c) sin θ
(d) cos θ

Ans: (a) cosec θ

Sol: cos² θ + sin² θsin θ = 1sin θ = cosec θ

Q3: Evaluate cos 60° sin 30° + sin 60° cos 30° (1 Mark)
(a) 1
(b) 3
(c) 1/2
(d) 3/2

Ans: (a) 1, 

Q4: If cos A = 2/5,  find the value of 4 + 4 tan²A
(a) 5
(b) 1/25
(c) 25
(d) 1/5

Ans: (d) 25

Q5: What is the value of (cos² 67° – sin² 23°) (1 Mark)
(a) 2
(b) 0
(c) 6
(d) 1

Ans: (b) 0
67° + 23° = 90°
∴ cos² 67° – sin² 23° = cos² (90° – 23°) – sin²23°
= sin² 23° – sin² 23°
= 0

Q6: Find the value of sin 38° – cos 52°? (2 Mark)

Ans:
sin 38° – cos 52°
= sin38° – cos (90° – 38°) [∵ cos (90° – θ = sin θ]
= sin 38° – sin 38°
= 0

Q7: Prove the following : (2 Mark)
 

Ans:  

 Here 1cos θ = sec θ and 1sin θ = cosec θ

LHS = 2cos² θ - 1cos⁴ θ - 2sin² θ + 1sin⁴ θ

= 2 sec² θ - sec⁴ θ - 2 cosec² θ + cosec⁴ θ

= 2(1 + tan² θ) - (1 + tan² θ)² - 2(1 + cot² θ) + (1 + cot² θ)²

= (1 + tan² θ)(2 - (1 + tan² θ)) - (1 + cot² θ)(2 - (1 + cot² θ))

= (1 + tan² θ)(1 - tan² θ) - (1 + cot² θ)(1 - cot² θ)

= 1 - tan⁴ θ - (1 - cot⁴ θ) = cot⁴ θ - tan⁴ θ

Hence proved 

Q8: If 7sin²θ + 3cos²θ = 4, then find the value of tan θ. (2 Marks)

Ans:

7sin²θ + 3cos²θ = 4

Dividing both sides by cos²θ

7tan²θ + 3 = 4sec²θ

= 4(1 + tan²θ)

⇒ 3tan²θ = 1

Q9: When sec 4A = cosec (A – 20°), here 4A is an acute angle, find out the value of A. (3 Marks)

Ans: sec 4A = cosec (A – 20°)

We get that sec 4A = cosec (90° – 4A)

To find out the value of A, thus, substitute the above equation in a given problem

cosec (90° – 4A) = cosec (A – 20°)

then, equate the angles

90° – 4A= A- 20°

110° = 5A

A = 110°/ 5 = 22°

Hence, the value of A = 22°

Q10: If 3x = sec θ and 9 x² - 1x² = tan θ, then find the value. (3 Marks)

Ans: 

9 x² - 1x² = 9 (sec θ)²3 - (tan θ)²3

= 9 sec² θ - tan² θ3

= 9 × 13 = sec² θ - tan² θ

= 1

Q11:  If ∠A and ∠B are the acute angles such that cos A = cos B, then show that ∠ A = ∠ B. (3 Marks)

Ans: Let us assume that the triangle ABC in which CD⊥AB

Given that the angles A and B are acute angles, such that

Cos (A) = cos (B)

As per the angles taken, the cost ratio is written as

AD/AC = BD/BC

Now, interchange the terms, we get

AD/BD = AC/BC

Let take a constant value, AD/BD = AC/BC = k

then consider the equation as

AD = k BD …(1)

AC = k BC …(2)

By applying the Pythagoras theorem in △CAD and △CBD we get,

CD² = BC² – BD² … (3)

CD² =AC² −AD² ….(4)

From the equations (3) and (4) we observe,

AC² − AD² = BC² − BD²

then substitute the equations (1) and (2) in (3) and (4)

k²(BC²−BD²) = (BC²−BD²) k² = 1

Putting this value in equation, we obtain that

AC = BC

∠A=∠B (Angles opposite to the equal side are equal-isosceles triangle)

Q12: In triangle ABC, right-angled at B, when tan A = 1/√3 find out the value : (5 marks)
(i) sin A cos C + cos A sin C
(ii) cos A cos C – sin A sin C

Ans:

Let’s ΔABC in which ∠B=90°

tan A = BC/AB = 1/√3

Let’s BC = 1k and AB = √3 k,

here k is the positive real number of the problem

By the Pythagoras theorem in ΔABC we get:

AC² = AB² + BC²

AC² = (√3 k)² + (k)²

AC² = 3k² + k²

AC² = 4k²

AC = 2k

then find the values of cos A, Sin A

Sin A = BC/AC = 1/2

Cos A = AB/AC = √3/2

now, find the values of cos C and sin C

Sin C = AB/AC = √3/2

Cos C = BC/AC = 1/2

then, substitute the values in the given problems

(i) sin A cos C + cos A sin C 
= (1/2) ×(1/2 )+ √3/2 ×√3/2 
= 1/4 + 3/4 
= 1

(ii) cos A cos C – sin A sin C 
= (√3/2 )(1/2) – (1/2) (√3/2 ) 
= 0

 Q13: In ∆ ABC, the right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (5 Marks)
(i) sin A, cos A
(ii) sin C, cos C

Ans:

In the given triangle ABC, right angled at B = ∠B = 90°

Given: AB = 24 cm as well as BC = 7 cm

According to the Pythagoras Theorem,

In the right-angled triangle, the squares of the hypotenuse sides are equal to the addition of the squares for the other two sides.

On applying the Pythagoras theorem, we observe

AC² = AB² + BC²

AC² = (24)² + 7²

AC² = (576+49)

AC² = 625cm²

AC = √625 = 25

thus, AC = 25 cm

(i) To find Sin (A), Cos (A)

We get that sine (or) Sin function is equal to the ratio of length of the opposite sides to the hypotenuse sides. So it becomes

Sin (A) = Opposite side /Hypotenuse side = BC/AC = 7/25

Cosine or Cos function is same as the ratio of the length of the adjacent side to the hypotenuse side so it becomes,

Cos (A) = Adjacent side/Hypotenuse side = AB/AC = 24/25

(ii) To find out Sin (C), Cos (C)

Sin (C) = AB/AC = 24/25

Cos (C) = BC/AC = 7/25