$The sides of a triangle are a = 5, b = 6 a n d c = 8, show that:8 c o s A + 16 c o s B + 4 c o s C = 17.$

Open in App

Solution

$We have, a = 4, b = 6 a n d c = 8 c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c} = \frac{7}{8} c o s B = \frac{a^{2} + c^{2} - b^{2}}{2 a c} = \frac{11}{16} c o s C = \frac{a^{2} + b^{2} - c^{2}}{2 a b} = \frac{- 1}{4} 8 c o s A + 16 c o s B + 4 c o s C = 8 \times \frac{7}{8} + 16 \times \frac{11}{16} + 4 \times (- \frac{1}{4}) = 17$

Suggest Corrections

Similar questions

8 \cos A + 16 \cos B + 4 \cos C = 17

Δ A B C

A = (3, 2), B = (- 5, 6)

C = (8, 3)

8

15

17

The sides of a right-angled triangle are p, p+1 and 5. Find the area of triangle, if the longest side measures 5 cm.

Join BYJU'S Learning Program