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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
Solution
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
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