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Let G be the centroid of triangle ABC and the circumcircle of triangle AGC touches the side AB at A If angleGAC = (pi)/(3) and a = 3b, then sin C is equal to |
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Answer» `(3)/(4)` `(a//2)/(sin (pi//3)) = (AD)/(sin C) RARR sin C = (sqrt3)/(2a) sqrt(2B^(2) + 2c^(2) -a^(2))` `=(sqrt3)/(2a) sqrt(3B^(2)) = (3)/(2) (b)/(a) = (1)/(2)` |
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