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Derive Projection formula from Law of cosines - Mathematics

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प्रश्न

Derive Projection formula from Law of cosines

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उत्तर

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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Properties of Triangle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Trigonometry - Exercise 3.9 [पृष्ठ १४३]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 3 Trigonometry
Exercise 3.9 | Q 11. (ii) | पृष्ठ १४३

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