Proof of Projection Formulae

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.

 Properties of Triangles






11 and 12 Grade Math

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