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Question

In right triangle ABC, $B C = 7 c m, A C - A B = 1 c m$ and $∠ B = 90^{o}$ . The value of cos A + cos B+ cos C is
$\frac{1}{7}$
$\frac{32}{24}$
$\frac{31}{25}$
$\frac{25}{31}$

A

\frac{31}{25}

B

\frac{1}{7}

C

\frac{32}{24}

D

\frac{25}{31}

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Solution

Verified by Toppr

$A C^{2} = A B^{2} + 49$
$(A C - A B) (A C + A B) = 49$
$A C + A B = 49$
$c o s A + c o s B + c o s C = \frac{24}{25} + 0 + \frac{7}{25} = \frac{31}{25}$

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