Cos θ=frac{b}{√{a^{2}+b^{2}}} ; where, 0<θ<90 ; then sin θ=ldots ldots ldots ldots

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8. Introduction to Trigonometry

easy

$\cos \theta=\frac{b}{\sqrt{a^{2}+b^{2}}} ;$ where, $0<\theta<90 ;$ then $\sin \theta=\ldots \ldots \ldots \ldots$

A

$\frac{a}{\sqrt{a^{2}+b^{2}}}$

B

$\frac{a}{b}$

C

$\frac{b}{a}$

D

$\frac{a b}{\sqrt{a^{2}+b^{2}}}$

Solution

$\sin ^{2} \theta=1-\cos ^{2} \theta=1-\frac{b^{2}}{a^{2}+b^{2}}=\frac{\left(a^{2}+b^{2}\right)-b^{2}}{a^{2}+b^{2}}=\frac{a^{2}}{a^{2}+b^{2}}$

$\therefore \sin \theta=\frac{a}{\sqrt{a^{2}+b^{2}}}$

Standard 10

Mathematics

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