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Chapter 20

Trigonometrical Ratios — Assertion-Reason Type Questions

Class - 9 RS Aggarwal Mathematics Solutions



Assertion-Reason Questions

Question 1

Assertion (A): The value of sin 60° cos 30° + cos 60° sin 30° is 0.

Reason (R): sin 90° = 0 and sin 0° = 1.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

Given,

sin60cos30+cos60sin3032×32+12×1234+1444=1.\Rightarrow \sin 60^\circ \cos 30^\circ + \cos 60^\circ \sin 30^\circ\\[1em] \Rightarrow \dfrac{\sqrt{3}}{2} \times \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} \times \dfrac{1}{2}\\[1em] \Rightarrow \dfrac{3}{4} + \dfrac{1}{4}\\[1em] \Rightarrow \dfrac{4}{4}\\[1em] = 1.

Assertion (A) is false.

As,

sin 90° = 1 and sin 0° = 0.

So,

Reason (R) is false.

Hence, Option 4 is the correct option.

Question 2

Assertion (A): In a right angled △ABC, if ∠ABC = 90°, AB = 3 cm, BC = 4 cm, then sin A = cos C.

Reason (R): sinθcosθ\dfrac{\sin θ}{\cos θ} = tan θ and sin θ ×\times cos θ = cot θ.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

In a right angled △ABC, if ∠ABC = 90, AB = 3 cm, BC = 4 cm, then sin A = cos C. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

In a right angled △ABC,

∠ABC = 90°, AB = 3 cm, BC = 4 cm

By pythagoras theorem,

Hypotenuse2 = Perpendicular2 + Base2

AC2 = (BC)2 + (AB)2

AC2 = 42 + 32

AC2 = 16 + 9

AC2 = 25

AC = 25\sqrt{25}

AC = 5 cm.

sinA=perpendicularhypotenuse=BCAC=45cosC=basehypotenuse=BCAC=45sinA=cosC\Rightarrow \sin A = \dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{BC}{AC} = \dfrac{4}{5}\\[1em] \Rightarrow \cos C = \dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{BC}{AC} = \dfrac{4}{5}\\[1em] \Rightarrow \sin A = \cos C

Assertion (A) is true.

As, sinθcosθ\dfrac{\sin θ}{\cos θ} = tan θ

but sin θ ×\times cos θ is not equal to cot θ.

So, Reason (R) is false.

Hence, Option 1 is the correct option.

Question 3

Assertion (A): sin27cos63\dfrac{\sin 27^\circ}{\cos 63^\circ} = 1.

Reason (R): sin(90° - θ) = cos θ and cos(90° - θ) = sin θ.

  1. A is true, R is false.

  2. A is false, R is true.

  3. Both A and R are true.

  4. Both A and R are false.

Answer

sin27cos63=sin(9063)cos63=cos63cos63=1\Rightarrow \dfrac{\sin 27^\circ}{\cos 63^\circ}\\[1em] = \dfrac{\sin (90^\circ - 63^\circ)}{\cos 63^\circ}\\[1em] = \dfrac{\cos 63^\circ}{\cos 63^\circ}\\[1em] = 1

Assertion (A) is true.

As, sin(90° - θ) = cos θ

and

cos(90° - θ) = sin θ

So,

Reason (R) is true.

Hence, Option 3 is the correct option.

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