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}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{32}{24}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{31}{25}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{25}{31}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

A B C, B C = 7 c m

A C - A B = 1 c m

∠ B = 90^{o}

cos A + cos B + cos C

$Δ$ ABC is right-angled at B, with AB = 7 cm and AC - BC = 1 cm. Which of the following is correct?

△ A B C

B

B C = 7 c m, A C - A B = 1

cos A + sin A

△ A B C, a = 5, b = 4

cos (A - B) = \frac{31}{32}

C

Join BYJU'S Learning Program