Question

5.
In figure, triangle ABC is right angled at A. Q and R are points on line BC and P is a point such that QP||AC
and RP || AB. Find angle P. ​

Answer 1

Answer:

AC||QP

angle C=angle Q

RP||AB

angle B=angle R

A+B+C=P+Q+R

P=A

P=90

Mark as brainliest if I deserve it

Answer 2

Given: In figure Δ ABC is a right angled triangle at A. Q and R are the points on line BC and P is a point such that QP || AC and RP || AB  

To find: Angle P

Step-by-step explanation:

As given

∠A= 90°

Now

QP || AC and RB is transversal

Therefore ∠ACB= ∠PQR -----(i)

Similarly

RP || AB and RB as transversal

Therefore ∠ABC =∠ QRP ----(ii)

By angle sum property which states that the sum of angles of a triangle is 180° we get

In Δ ABC

$\angle A + \angle ABC +\angle ACB = 180^\circ ----(iii)$

and In Δ PQR

$\angle P + \angle QRP +\angle PQR = 180^\circ ----(iv)$

From (iii) and  (iv) we get

$\angle A + \angle ABC +\angle ACB =\angle P + \angle QRP +\angle PQR$

From (i) and (ii) we get

$\angle A + \angle QRP +\angle PQR =\angle P + \angle QRP +\angle PQR\\\\\Rightarrow \angle A = \angle P$

Hence the value of ∠P is 90°