Question

angle ABC=60.point P is inside it. such a that which is at a distant 2cm & 11cm from sides of angle ABC
then find length BP.

Answer 1

$\mathrm{PA} = 2 \mathrm{~cm}$

PC=11 cm

angle ABC=60°

BP-?

Denote angle ABP as a. Then angle PBC = ABC-ABP=60°-a.

Triangle ABP:

$\sin a = AP / BP$ (from definition), therefore $BP = AP / \sin (60{}^{\circ} - a)$ (1)

Triangle PBC:

$\sin (60{}^{\circ} - a) = PC / BP$ (from definition), therefore $BP = PC / \sin (60{}^{\circ} - a)$ (2)

From (1) and (2):

AP/sin a = PC/sin(60°-a) therefore PC/AP=sin(60°-a)/sin a (3)

From formula of angle difference for sine (sin a-b=sin a *cos b - sin b * cos a)

$\sin (60{}^{\circ} - a) = \sin 60{}^{\circ} * \cos a - \cos 60{}^{\circ} * \sin a = \frac{\sqrt{3}}{2} \cos a - \frac{1}{2} \sin a$ (4)

Substitute (4) into (3)

$\mathrm{PC} / \mathrm{AP} = \frac{\sqrt{3}}{2} (\cos a / \sin a) - \frac{1}{2} (\sin a / \sin a) = \frac{\sqrt{3}}{2} ctg a - \frac{1}{2} = 11 / 2$ using elementary math ctg $a = \frac{10}{\sqrt{3}}$

Triangle ABP:

ctg a=AB/AP

AB=AP*ctg a, using Pythagorean theorem

$$B P = \sqrt {A B ^ {2} + A P ^ {2}} = \sqrt {\frac {4 0 4}{3}}$$

Answer: $\mathrm{BP} = \sqrt{\frac{404}{3}}$