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secondary 3 | E Maths
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Date Posted: 3 years ago
Views: 264
By sine rule,
AC / sin B = AB / sin C = BC / sin A
So (BC / sin A)² = (AC / sin B)(AB / sin C)
BC² / sin² A = (AC)(AB) / (sin B sin C)
BC² = (AC)(AB) sin² A / (sin B sin C)
(You can skip this step if you are able to see that the numerator (AC)(AB) and denominator sin²A can be swapped immediately instead of performing any cross multiplication)
BC² / (AC)(AB) = sin² A / (sin B sin C)
BC² / (AC)(AB) = sin² (180° - A) / (sin B sin C)
(Recall sin A = sin (180° - A) )
BC² / (AC)(AB) = sin² (B + C) / (sin B sin C)
(Since A + B + C = 180°)
(Proved)
AC / sin B = AB / sin C = BC / sin A
So (BC / sin A)² = (AC / sin B)(AB / sin C)
BC² / sin² A = (AC)(AB) / (sin B sin C)
BC² = (AC)(AB) sin² A / (sin B sin C)
(You can skip this step if you are able to see that the numerator (AC)(AB) and denominator sin²A can be swapped immediately instead of performing any cross multiplication)
BC² / (AC)(AB) = sin² A / (sin B sin C)
BC² / (AC)(AB) = sin² (180° - A) / (sin B sin C)
(Recall sin A = sin (180° - A) )
BC² / (AC)(AB) = sin² (B + C) / (sin B sin C)
(Since A + B + C = 180°)
(Proved)
Edited
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Date Posted:
3 years ago
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