Question

From the given figure; find:
(i) tan x°
(ii) sin y°
(iii) cos y°​

Answer 1

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(i) In right angled △SRB,

$\sf \: tan \: x \degree = \frac{p}{b} =\frac {RB}{RS} = \frac{6}{5}$

(ii) In right angled △ARS,

AR=18-6=12

$\sf \: AS^2=AR^2+SR^2$

$\rightarrow \sf {12}^{2} + {5}^{2} \\ \\ \rightarrow \sf144 + 25 = 169 \\ \\ \sf \: AS = \sqrt{169} = 13$

$\therefore \sf \: sin \: y \degree = \frac{p}{h} = \frac {SR}{AS}= \frac{5}{13}$

$\sf \: (iii) \: cos \: y°=\frac {b}{h}=\frac{AR}{AS} = \frac{12}{13}$

Hope it helps you..

Answer 2

Answer:

In triangle SRB

Tanx°=opposite side /adjacent side

=BR/SR

=6/5

AR=AB-BR

=18-6

=12

In triangle SAR

by Pythagoras theorem

AS^2=SR^2+AR^2

=5^2+12^2

=25+144

=169

AS=13

Sin y°=opposite side/ hypoteneus

=SR/SA

=5/13

cos y°=adjacent side/hypoteneus

=AR/AS

=12/13