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# The Projection Law

From the Sine Law, We have,
b =2R sin B...........(i)
c =2R sin C...........(ii)
=2R sin A...........(iii)

To Prove: i) cos C + cos B =
Taking LHS;
= b cos C + cos B
= 2R sin B. cos C + 2R cos B. sin C  [from eqn (i) & (ii)]
=2R(sin B. cos C + cos B. sin C )
= 2R sin(B + C)
= 2R sin A                                              [B + C=180O ﻿- A, so, sin(B + C)=sin(180O ﻿- A)= sin A]
=R.H.S.                                          [from eqn (iii)]

To Prove: ii) c cos A +  cos C = b
Taking LHS;
= c cos A +  cos C
= 2R sin C. cos A + 2R cos C. sin A  [from eqn (ii) & (iii)]
= 2R(sin C. cos A + cos C. sin A)
= 2R sin(C + A)
= 2R sin B                                              [ C + A=180O ﻿- B, so, sin(C + A)=sin(180O ﻿- B)= sin B]
= R.H.S.                                           [from eqn (i)]

To Prove: iii) cos B + b cos A = c
Taking LHS;
cos B + cos A
= 2R sin A. cos B + 2R cos A. sin B  [from eqn (iii) & (i)]
= 2R(sin A. cos B + cos A. sin B)
= 2R sin(A + B)
= 2R sin C                                              [ A + B=180O ﻿- C, so, sin(A + B)=sin(180O ﻿- C)= sin C]
= c = R.H.S.                                           [from eqn (ii)]