For the given figure Find ${sin}^{2} C + {cos}^{2} C$

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Solution

The correct option is A $1$
In $△ A B C$ , $A D ⊥ B C$ $A B = 13$ , $B D = 5$ , $D C = 16$
Now, In $△ A B D$ ,
$A B^{2} = A D^{2} + B D^{2}$
$13^{2} = A D^{2} + 5^{2}$
$A D^{2} = 144$
$A D = 12$
Now, in $△ A D C$ ,
$A C^{2} = A D^{2} + C D^{2}$
$A C^{2} = 12^{2} + 16^{2}$
$A C^{2} = 144 + 256$
$A C = 20$ cm
Now, ${sin}^{2} C + {cos}^{2} C = (\frac{P}{H})^{2} + (\frac{B}{H})^{2} = (\frac{A D}{A C})^{2} + (\frac{B D}{A C})^{2}$
= $(\frac{12}{20})^{2} + (\frac{16}{20})^{2}$
= $\frac{144}{400} + \frac{256}{25}$
= $\frac{400}{400}$
= $1$

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