Mathematics

In $\Delta ABC, \angle C=90^{\circ}$ then the value of $\sin^{2}A+\sin^{2}B$ is :

$1$

SOLUTION
Given $∠C={90}^o$
From Triangle,
$\sin A =\cfrac {BC}{AB};$  $\sin B=\cfrac {AC}{AB}$
$\Rightarrow {\sin}^2A+{\sin}^2B$
$\cfrac { { BC }^{ 2 } }{ { AB }^{ 2 } } +\cfrac { { AC }^{ 2 } }{ { AB }^{ 2 } } =\cfrac { { BC }^{ 2 }+{ AC }^{ 2 } }{ { AB }^{ 2 } }$
From $\triangle ABC$, by using Pythagoras Theorem,
${AB}^2={BC}^2+{AC}^2$
${\sin}^2A+{\sin}^2B=\cfrac { { BC }^{ 2 }+{ AC }^{ 2 } }{ { AB }^{ 2 } }=\cfrac { { AB }^{ 2 } }{ { AB }^{ 2 } }=1$ You're just one step away

Single Correct Medium Published on 09th 09, 2020
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