In triangle ABC, B, sin{A}=\frac{7}{25}, then the value of \cos{C}is

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Solution

Q.
In $∆ A B C$ , right angled at B $s i n A = \frac{7}{25}$ , then the value of cos C is

1 $\frac{7}{25}$

2 $\frac{24}{25}$

3 $\frac{7}{24}$

4 $\frac{24}{7}$

Ans.
(1)

$s i n A = \frac{p}{h} = \frac{B C}{A C} = \frac{7}{25} c o s C = \frac{b}{h} = \frac{B C}{A C} = \frac{7}{25}$

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