Question

In triangle ABC BP and CP are the bisector of angle abc and angle ACB respectively find angle BAC and the exterior angle of point P is 340°

Answer 1

Given : In triangle ABC BP and CP are the bisector of angle abc and angle ACB respectively the exterior angle of point P is 340°

To Find :  angle BAC

Solution:

∠ABC  = 2x

∠ACB = 2y

∠ABC + ∠ACB  + ∠BAC = 180°   ( sum of angles of a triangle )

=> 2x + 2y + ∠BAC = 180°

=> ∠BAC = 180 - 2x - 2y

=> ∠BAC = 180 - (2x+ 2y )

in ΔBCP

∠BCP = 360° - 340° = 20°

∠PBC  = ∠ABC /2  = 2x/2 = x

∠PCB  = ∠ACB /2  = 2y/2 = y

=> x  + y  + 20 = 180

=> x + y = 160

=> 2x + 2y  = 320

∠BAC = 180  - 320 = -140°

angle should not be in negative

Hence there is mistake in data

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Answer 2

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1) In ΔABC, ∠ABC = 2∠ACB

Let ∠ACB = x

⇒∠ABC = 2∠ACB = 2x

Given BP is bisector of ∠ABC

Hence ∠ABP = ∠PBC = x

Using the angle bisector theorem, that is,

the bisector of an angle divides the side opposite to it in the ratio of other two sides.

Hence, CB : BA= CP:PA.

2) Consider ΔABC and ΔAPB

∠ABC = ∠APB [Exterior angle property]

∠BCP = ∠ABP [Given]

∴ ΔABC ≈ ΔAPB [AA criterion]

∴fraction numerator space AB over denominator BP end fraction space equals space CA over CB[Corresponding sides of similar triangles are proportional.]

⇒ AB x BC = BP x CA