The geometrical interpretation of the proof of projection formulae is the length of any side of a triangle is equal to the algebraic sum of the projections of other sides upon it.
In any triangle ABC,
(i) a = b cos C + c cos B
(ii) b = c cos A + a cos C
(iii) c = a cos B + b cos A
Proof:
In any triangle ABC we have a
\(\frac{a}{sin A}\) = \(\frac{b}{sin B}\) = \(\frac{c}{sin C}\) = 2R ……………………. (1)
Now convert the above relation into sides in terms of angles
in terms of the sides of any triangle.
a/sin A = 2R
⇒ a = 2R sin A ……………………. (2)
b/sin B = 2R
⇒ b = 2R sin B ……………………. (3)
c/sin c = 2R
⇒ c = 2R sin C ……………………. (4)
(i) a = b cos C + c cos B
Now, b cos C + c cos B
= 2R sin B cos C + 2R sin C cos B
= 2R sin (B + C)
= 2R sin (π - A), [Since, A + B + C = π]
= 2R sin A
= a [From (2)]
Therefore, a = b cos C + c cos B. Proved.
(ii) b = c cos A + a cos C
Now, c cos A + a cos C
= 2R sin C cos A + 2R sin A cos C
= 2R sin (A + C)
= 2R sin (π - B), [Since, A + B + C = π]
= 2R sin B
= b [From (3)]
Therefore, b = c cos A + a cos C.
Therefore, a = b cos C + c cos B. Proved.
(iii) c = a cos B + b cos A
Now, a cos B + b cos A
= 2R sin A cos B + 2R sin B cos A
= 2R sin (A + B)
= 2R sin (π - C), [Since, A + B + C = π]
= 2R sin C
= c [From (4)]
Therefore, c = a cos B + b cos A.
Therefore, a = b cos C + c cos B. Proved.
11 and 12 Grade Math
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