# ML Aggarwal Solutions Class 9 Mathematics Solutions for Trigonometric Ratios Exercise 17 in Chapter 17 - Trigonometric Ratios

Question 5 Trigonometric Ratios Exercise 17

In triangle ABC, AB = 15 cm, AC = 15 cm and BC = 18 cm.

Answer:

Here ABC is a triangle in which

AB = 15 cm, AC = 15 cm and BC = 18 cm

Draw AD perpendicular to BC, D is mid-point of BC.

Then, BD – DC = 9 cm

in right-angled triangle ABD,

By Pythagoras theorem, we get

\begin{array}{l} A B^{2}=A D^{2}+B D^{2} \\ A D^{2}=A B^{2}-B D^{2} \end{array}

\begin{array}{l} A D^{2}=(15)^{2}-(9)^{2} \\ A D^{2}=225-81 \\ A D^{2}=144 \\ A D-12 \mathrm{cm} \end{array}

(i) cos ∠ABC = Base// Hypotenuse

(In right angled ∆ABD, ∠ABC = ∠ABD)

= BD / AB

= 9/15

= 3/5

(ii) sin ∠ACB = sin ∠ACD

= perpendicular/ Hypotenuse

= AD/AC

= 12/15

= 4/5

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