HOTS & Value Based Questions: Introduction to Trigonometry

# Class 10 Maths Chapter 8 HOTS Questions - Introduction to Trigonometry

## HOTS

Q1: What is the maximum value of

Ans:  ∵    = 3 sin θ
Since the maximum value of ‘sin θ’ is 1
∴  the maximum value of 3 sin θ is 3 × 1 i.e. 3

⇒  the maximum value of  is 3.

Q2: If sin θ = 1/3,  then find the value of 9 cot2θ + 9.

Ans: Thus, the value is 81.

Q3: If 4 tan θ = 3, then find the value of

Ans:

Given $4 \tan \theta = 3$4tanθ=3, we have:

Let sinθ=3k and cosθ=4k (since $\tan \theta = \frac{\sin \theta}{\cos \theta}$tanθ = sinθ/cosθ). Using the identity sin2θ + cos2θ=1:

(3k)+ (4k)= $19k^2 + 16k^2 = 1$

9k+ 16k= 1

25k= 1

⇒k= 1/5

So, sinθ = 3/5 and $\cos \theta = \frac{4}{5}$cosθ = 4/5. Now calculate the expression: $\tan \theta = \frac{3}{$

Q4: If sin α = 1/2 and cos β = 1/2 then find the value of (α + β).

Ans:

Q5: If sin θ + cos θ = √3 , find the value of tan θ + cot θ .

Ans: sin θ + cos θ = 3 ⇒ (sin θ + cos θ)2 = 3
⇒ sin2 θ + cos2 θ + 2 sin θ . cos θ = 3
⇒ 1 + 2 sin θ . cos θ = 3 ⇒ 2 sin θ . cos θ = 2    [∴ sin2θ + cos2θ = 1]

⇒ sin θ . cos θ =1 ⇒  1 =

⇒

Thus, tanθ + cotθ = 1

Q6: cos (A+B) = 1/2 and sin (A–B) = 1/2 ; 0° < (A + B) < 90° and (A – B) > 0°. What are the values of ∠A and ∠B?

Ans:

...(1)

...(2)

Adding (1) and (2), 2A = 90 ⇒ A = 45
From (1) 45° + B = 60°  ⇒   B = 60° – 45° = 15°
Thus, ∠A = 45° and ∠B = 15°

Q7: Prove that sec2A + cosec2A = sec2A. csec2 A
Ans:

= sec2A⋅csec2A

Thus, the equation is proved.$\sec^2 A + \csc^2 A = \frac{1}{\cos^2 A} + \frac{1}{\sin^2 A}$

Q8: Prove that (sinθ + cosecθ)2 + (cosθ + secθ)2 = tan2θ + cot2θ + 7
Ans:

LHS =  sin2θ + cosec2θ + 2sinθcosecθ + cos2θ + sec2θ + 2cosθsecθ
1 + 1 + cot2θ + 1 + tan2θ + 2 + 2
= 7 + tan2θ+cot2θ

Q9: If cos(40o + x) = sin 30o, find the value of x, provided 40o + x is an acute angle.
Ans: Given that
cos(40o + x) = sin 30o
Now RHS = sin 30o = 1/2
So, cos (40o + x)  = 1/2
We know that cos60o  = 1/2, therefore
40 + x = 60o
x = 20o

Q10: Prove that .
Ans:

LHS =

## Value-based Questions

Q1: A group of students plan to put up a banner in favour of respect towards girls and women, against a wall. They placed a ladder against the wall such that its foot is at a distance of 2.5 m from the wall and its top reaches a window 6 m above the ground.

(i) Find the length of the ladder.
(ii) Which mathematical concept is used in this problem?
(iii) By putting up a banner in favour of respect to women, which value is depicted by the group of students?

Ans: (i) Let AB be the ladder and CA be the wall with the window at A. Also,
BC = 2.5 m
CA = 6m
∴ In rt ΔACB, we have
AB2 = BC2 + CA2 [Using Pythagoras Theorem]
= (2.5)2 + (6)2
= 6.25 + 36 = 42.25

⇒

Thus, the length of the ladder is 6.5 m.
(ii) Triangles (Pythagoras Theorem)
(iii) Creating positive awareness in public regarding women.

Q2: Raj is an electrician in a village. One day power was not there in entire village and villagers called Raj to repair the fault. After thorough inspection he found an electric fault in one of the electric pole of height 5 m and he has to repair it. He needs to reach a point 1.3m below the top of the pole to undertake the repair work. On the basis of above, answer the following question.

(i) When the ladder is inclined at an angle of α such that 3tan α + 2 = 5 to the horizontal, find the angle $\al$α.
(ii) How far from the foot of the pole should he place the foot of the ladder? (Use $\sqrt{3} = 1.73$3=1.73)
(iii) In the above situation, find the value of .
(iv) In the above situation, if BD=3 cm and BC=6 cm, find α.
(v) In your opinion, how does Raj's action reflect the values of responsibility and community service?

Ans:
(i) Given the equation 3tanα+2=5, simplify to find $\tan \alpha$tanα:We know that tan60º=3, so the angle α=60º.

(ii) Raj needs to reach a height of 3.7m (since $5−1.3=3.7 m5 - 1.3 = 3.7 \, \text{m}$) and the ladder is inclined at 60º

Using tan60º = 3.7/x and knowing tan60º = 1.73, we get:

Thus, the distance from the foot of the pole to the foot of the ladder is 2.14 m.

(iii) The expressionis the sine subtraction identity:
Simplifying Thus, the expression becomes:

Given α=60º, we have:

(iv) We are given BD=3 cm and BC=6 cm. Using the ratio of sides, we find that
=
This corresponds to $tan⁡30$º, so $\alpha = 30^\circ$α=30º.

(v)  Raj's actions reflect a strong sense of responsibility because he promptly responded to the village's call for help, even in challenging conditions. His commitment to restoring power shows that he cares about the welfare of others. This sense of responsibility is a key value in community service, where individual efforts contribute to the well-being of the whole society.

$\cos 60^\circ = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{5}$

The document Class 10 Maths Chapter 8 HOTS Questions - Introduction to Trigonometry is a part of the Class 10 Course Mathematics (Maths) Class 10.
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## Mathematics (Maths) Class 10

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## FAQs on Class 10 Maths Chapter 8 HOTS Questions - Introduction to Trigonometry

 1. What are the basic trigonometric ratios and how are they defined?
Ans.The basic trigonometric ratios are sine (sin), cosine (cos), and tangent (tan). They are defined based on the sides of a right-angled triangle. For an angle θ: - sin(θ) = opposite side / hypotenuse - cos(θ) = adjacent side / hypotenuse - tan(θ) = opposite side / adjacent side.
 2. How can trigonometry be applied in real-life situations?
Ans.Trigonometry is applied in various real-life situations, such as in architecture to calculate heights and distances, in navigation for determining positions, and in physics for analyzing wave patterns. It is also used in computer graphics, engineering, and even in music theory.
 3. What is the Pythagorean theorem and how does it relate to trigonometry?
Ans.The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse (c) is equal to the sum of the squares of the other two sides (a and b). It is expressed as a² + b² = c². This theorem is fundamental in trigonometry because it helps derive the trigonometric ratios and proves their relationships.
 4. What are some common identities in trigonometry?
Ans.Some common trigonometric identities include: - Pythagorean identities: sin²(θ) + cos²(θ) = 1 - Angle sum identities: sin(A + B) = sin(A)cos(B) + cos(A)sin(B) - Double angle identities: sin(2θ) = 2sin(θ)cos(θ). These identities are useful for simplifying expressions and solving equations.
 5. How do you solve basic trigonometric equations?
Ans.To solve basic trigonometric equations, you typically isolate the trigonometric function on one side of the equation. For example, if you have sin(θ) = 0.5, you can find θ by taking the inverse sine (arcsin). Additionally, consider the periodic nature of trigonometric functions to find all possible solutions in the given interval.

## Mathematics (Maths) Class 10

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