## ML Aggarwal Class 9 Solutions for ICSE Maths Chapter 17 Trigonometric Ratios Chapter Test

Question 1.

(a) From the figure (i) given below, calculate all the six t-ratio for both acute angles.

(b) From the figure (ii) given below, find the values of x and y in terms of t-ratio of θ.

Answer:

Given ∆ ABC is right angled at B,

Then in right angled ∆ ABC,

By Pythagoras theorem, we get

AC^{2} = AB^{2} + BC^{2}

⇒ AB^{2} = AC^{2} – BC^{2}

⇒ AB^{2} = (3)^{2} – (2)^{2}

⇒ AB^{2} = 9 – 4

⇒ AB^{2} = 5

AB = \( \sqrt{{5}} \)

(b) Let ABC be given triangle in which B is at right angle and ∠ BAC = θ

Then we know that,

Question 2.

(a) From the figure (1) given below, find the values of:

(i) sin ∠ ABC

(ii) tanx – cosx + 3sinx.

(b) From the figure (2) given below, find the values of;

(i) 5 sin x

(ii) 7 tan x

(iii) 5 cos x – 17 siny – tan x.

Answer:

(a) In this figure

BC = 12, CD = 9 and BC = 20

In right angled ∆ ABC,

By Pythagoras theorem, we get

AB^{2} = AC^{2} + BC^{2}

⇒ ^{2} = AB^{2} – BC^{2}

⇒ AC^{2} = (20)^{2} – (12)^{2}

⇒ AC^{2} = 400 – 144

⇒ AC^{2} = 256 ⇒ AC^{2} = (16)^{2} ⇒ AC = 16

But, In right angled ∆ BCD,

By Pythagoras theorem, we get

BD^{2} = ^{2} + CD^{2}

⇒ BD^{2} = (12)^{2} + (9)^{2}

BD^{2}= 144 + 81 ⇒ BD^{2} = 225 ⇒ BD^{2} = (15)^{2}

⇒ BD = 15

(b) From figure AC = 17, AB = 25, AD = 15

In right angled ∆ ACD By Pythagoras theorem, we get

AC^{2} = AD^{2} + CD(17)^{2} = (15)^{2} + (CD)^{2}

⇒ CD^{2} = (17)^{2} – (15)^{2}

⇒ CD^{2} = 289 – 225

⇒ CD^{2} = 64

⇒ CD^{2} = (8)^{2}

⇒ CD = 8

In right angled ∆ABD,

By Pythagoras theorem, we get ,

AB^{2} = AD^{2} + BD^{2}

⇒ (25)^{2} = (15)^{2} + BD^{2} ⇒ BD^{2} = (25)^{2} – (15)^{2}

⇒ BD^{2} = 625 – 225 ⇒ BD^{2} = 400

⇒ BD^{2} = (20)^{2} ⇒ BD = 20

Question 3.

If q cosθ = p, find tan θ – cot θ in terms of p and q.

Answer:

Let ABC be a right angled triangle and B at right angled and ∠ ACB = 0 .

Given that,

q cos θ = p ⇒ cos θ = \(\frac{p}{q}\)

⇒ \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{p}{q}\)

Let BC = px then AC = qx

In right angled ∆ ABC,

By Pythagoras theorem, we get

AC^{2} = AB^{2} + BC^{2}

⇒ AB^{2} = AC^{2} – BC^{2}

⇒ AB^{2} = (qx)^{2} – (px)^{2}

⇒ AB^{2} = q^{2}x^{2} – p^{2}x^{2}

⇒ AB^{2} = (q^{2} – p^{2}) x^{2}

Question 4.

Given 4 sin θ =3 cos θ, find the values of :

(i) sin θ

(ii) cos θ

(iii) cot^{2} θ – cosec^{2} θ.

Answer:

Given that, 4sin θ = 3cosθ

⇒ \(\frac{\sin \theta}{\cos \theta}=\frac{3}{4}\)

Question 5.

If 2 cos θ = \( \sqrt{{3}} \), prove that 3 sin θ – 4 sin^{3} θ = 1

Answer:

2 cos θ = \( \sqrt{{3}} \)

Hence proved.

Question 6.

If \(\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}=\frac{1}{4}, \text { find } \sin \theta\)

Answer:

Question 7.

If sin θ + cosec θ = \(3 \frac{1}{3}\), find the value of sin^{2} θ + cosec^{2} θ.

Answer:

Question 8.

In the adjoining figure, AB = 4 m and ED = 3 m.

If sin α = \(\frac{3}{5}\) and cos β = \(\frac{12}{13}\) , And the length of BD.

Answer:

sin α = \(\frac{3}{5}\) = \(\frac{AB}{AC}\)

∴ AB = 3, AC = 5

But AC^{2} = AB^{2} + BC^{2}

⇒ (5)^{2} = (3)^{2} + BC^{2} ⇒ 25 = 9 + BC^{2}

⇒ BC^{2} = 25 – 9 = 16 = (4)^{-2} ⇒ BC = 4

∴ tan α = \(\frac{AB}{BC}\) = \(\frac{4}{5}\)

cos β = \(\frac{12}{13}\) = \(\frac{CD}{CE}\)

∴ CD = 12, CE = 13

But CE^{2} = CD^{2} + ED^{2}

(13)^{2} = (12)^{2} + (ED)^{2}

⇒ 169 = 144 + ED^{2} ⇒ ED^{2} = 169 – 144 = 25 = (5)^{2}

∴ ED = 5