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Derive Projection formula from Law of sines - Mathematics

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प्रश्न

Derive Projection formula from Law of sines

योग

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 the Law of sines,

We know that `"a"/sin"A" = "b"/sin "B" = "c"/sin "C"` = 2R

`"a"/sin"A"` = 2R ⇒ a = 2R sin A

`"b"/sin"B"` = 2R ⇒ b = 2R sin B

`"c"/sin"C"` = 2R ⇒ c = 2R sin C

(a) b cos C + c cos B = 2 R sin B cos C + 2 R sin C cos B

= 2R (sin B cos C + cos B sin C)

= 2R sin (B + C)

= 2R sin (180° – A)

b cos C + c cos B = 2R sin A = a

a = b cos C + c cos B

(b) c cos A + a cos C = 2R sin C cos A + 2R sin A cos C

= 2R (sin C cos A + cos C sin A)

= 2R sin(C + A)

= 2R sin(180° – B)

= 2R sin B = b

∴ b = c cos A + a cos C

(c) a cos B + b cosA = 2R sin A cos B + 2R sin B cos A

= 2R (sin A cos B + cos A sin B)

= 2R sin(A + B)

= 2R sin(180° – C)

= 2R sin 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. (i) | पृष्ठ १४३

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