Question

hello everyone!!☺☺☺

if sin theta = 8 by 9
find sin theta + cos theta ???

Answer 1

$\sin \theta = \frac{8}{9} = \frac{p}{h} \\ b = \sqrt{ {9}^{2} - {8}^{2} } = \sqrt{81 - 64} = \sqrt{17} \\ \cos\theta = \frac{b}{h} = \frac{ \sqrt{17} }{9} \\ \sin\theta + \cos\theta = \frac{8}{9} + \frac{ \sqrt{17} }{9} \\ = \frac{8 + \sqrt{17} }{9}$

Answer 2

sin theta =8/9 , =p/h

therefore p=8 & h=9

using Pythagoras theorem b=√h² - p²

=√81 - 72

= √9

=3

now,

cos theta = b/h

=3/9

ATQ

sin theta + cos theta

=8/9 + 3/9

=11/9