Q. If a right triangle ABC, …
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Sia ? 6 years, 6 months ago
In {tex} \triangle A B C{/tex},
{tex}\tan A = 1{/tex}
{tex}\Rightarrow \quad \frac { B C } { A C } = 1{/tex}
{tex}\Rightarrow {/tex} BC = x and AC = x
Using Pythagoras theorem,
{tex}\Rightarrow A B ^ { 2 } = A C ^ { 2 } + B C ^ { 2 }{/tex}
{tex}\Rightarrow \quad A B ^ { 2 } = x ^ { 2 } + x ^ { 2 }{/tex}
{tex}\Rightarrow \quad A B = \sqrt { 2 } x{/tex}
{tex}\therefore \quad \sin A = \frac { B C } { A B } = \frac { x } { \sqrt { 2 } x } = \frac { 1 } { \sqrt { 2 } } \text { and } \cos A = \frac { A C } { \sqrt { 2 } x } = \frac { x } { \sqrt { 2 } x } = \frac { 1 } { \sqrt { 2 } } {/tex}
2 sin A cos A {tex}= 2 \times \frac { 1 } { \sqrt { 2 } } \times \frac { 1 } { \sqrt { 2 } } = 1{/tex}
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