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Question
Derive Projection formula from Law of cosines
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Solution
To prove
(a) a = b cos C + c cos B
(b) b = c cos A + a cos C
(c) c = a cos B + b cos A
Using Law of cosines.
We have cos A = `("b"^2 + "c"^2 - "a"^2)/(2"bc")`
cos B = `("c"^2 + "a"^2 - "b"^2)/(2"ca")`
cos C = `("a"^2 + "b"^2 - "c"^2)/(2"ab")`
(a) b cos C + cs B
= `"b"(("a"^2 + "b"^2 - "c"^2)/(2"ab")) + "c"(("c"^2 + "a"^2 - "b"^2)/(2"ca"))`
= `("a"^2 + "b"^2 - "c"^2)/(2"a") + ("c"^2 + "a"^2 - "b"^2)/(2"a")`
= `("a" + "b"^2 - "c"^2 + "c"^2 + "a"^2 - "b"^2)/(2"a")`
= `(2"a"^2)/(2"a")`
= a
∴ a = b cos C + c cos B
(b) c cos A + a cos C
= `"c"(("b"^2 + "c"^2 - "a"^2)/(2"bc")) + "a"(("a"^2 + "b"^2 - "c"^2)/(2"ab"))`
= `("b"^2 + "c"^2 - "a"^2)/(2"bc") + ("a"^2 + "b"^2 - "c"^2)/(2"ab")`
= `("b"^2 + "c"^2 - "a"^2 + "a"^2 + "b"^2 - "c"^2)/(2"b")`
= `(2"b"^2)/(2"b")`
= b
∴ b = c cos A + a cos C
(c) c = a cos B + b cos A
= `"a"(("c"^2 + "a"^2 - "b"^2)/(2"ca")) + "b"(("b"^2 + "c"^2 - "a"^2)/(2"bc"))`
= `("c"^2 + "a"^2 - "b"^2)/(2"c") + ("b"^2 + "c"^2 - "a"^2)/(2"c")`
= `("c"^2 + "a"^2 - "b"^2 + "b"^2 + "c"^2 - "a"^2)/(2"c")`
= `(2"c"^2)/(2"c")`
= c
∴ c = a cos B + b cos A
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