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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