Question

73. In a triangle ABC, BP is drawn perpendicular to BC to meet CA in P, such that CA = AP, then
BP
AB
1) 2 sin A
2) 2 sin B
3) 2 sin C
4) none of these​

Answer 1

Answer:

In triangle ABC, BP is drawn perpendicular to BC to meet CA produced in P such that CA=AP. Then what is BP/AP?

Triangle ABC is a right-angled triangle with AP as the hypotenuse, BC as the base,b, and BP as the altitude, h. A is center of the circumscribing circle.

CP = hypotenuse = [b^2+h^2]^0.5 and AP = AC = CP/2 = [b^2+h^2]^0.5/2

Ratio BP/AP = h/{[b^2+h^2]^0.5/2} = 2h/[b^2+h^2]^0.5 = 2 sin ACB.

Answer 2

Answer:

2 sin c

Step-by-step explanation:

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