Question

If the sides of a ∆ABC are a = 4, b = 6 and c = 8, then show that 4 cos B + 3 cos C = 2

Answer 1

a = 4,

b = 6,

c = 8

To prove 4 cos B + 3 cos C = 2

cos B = `("a"^2 + "c"^2 - "b"^2)/(2"ac")`

= `(16 + 64 - 36)/(2(4)(8))`

= `(80 - 36)/64`

= `44/64`

= `11/16`

cos C = `("a"^2 + "b"^2 - "c"^2)/(2"ab")`

= `(16 + 36 - 64)/(2(4)(6))`

= `(-12)/48`

= `- 1/4`

So L.H.S = 4 cos B + 3 cos C

= `4(11/16) + 3(- 1/4)`

= `11/4 - 3/4`

= `8/4`

= 2

= R.H.S