The length of CP is 4. What is the length of AB?
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∠ C B A = 6 0 tan ( ∠ C B A ) = tan ( 6 0 ) = 3 A B A C = 3 A B 4 + A P = 3 3 4 + A P = A B . . . ( i ) ∠ P B A = 3 0 tan ( 3 0 ) = 3 1 A B A P = 3 A B = 3 A P . . . ( i i ) (i) and (ii) ⟹ 3 4 + A P = 3 A P 4 + A P = 3 A P A P = 2 ∴ A B = 3 A P = 3 2 = 2 3
The image shows half of an equilateral triangle with point P its incenter. That makes A P = 4 / 2 = 2 , A B = 2 / tan ( 3 0 ) = 2 3 .
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BP = CP = 4; AB = BP*cos30