Question : $\triangle ABC$ is a right angled triangle, $\angle B$ = 90°, $AB$ = 12 cm, $BC$ = 5 cm. What is the value of $\cos A + \sin C$?
Option 1: $\frac{24}{13}$
Option 2: $\frac{25}{13}$
Option 3: $\frac{10}{13}$
Option 4: $\frac{12}{13}$
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Correct Answer: $\frac{24}{13}$
Solution :
In $\triangle$ABC,
⇒ $AC$
2
= $AB$
2
+ $BC$
2
⇒ $AC$
2
= 12
2
+ 5
2
⇒ $AC$ = 13 cm
$\cos A =\frac{\text{Base}}{\text{Hypotenuse}}=\frac{12}{13} $
$\sin C =\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{12}{13} $
Therefore, the value of cos A + sin C,
$\therefore \cos A + \sin C =\frac{12}{13}$ + $\frac{12}{13}=\frac{24}{13} $
Hence, the correct answer is $ \frac{24}{13}$.
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