Question

please help me with the question below;

(In triangle PQR angle R=100° and line segment RS bisects angle R such that RS is perpendicular to PQ find angle P?)​

Answer 1

Answer:

|R = 100° , RS is perpendicular to PQ,

to find |P

in ∆ PQR, |PRS = 50° [ RS bisects |R , given ]

and |PSR = 90° [ RS is perpendicular to PQ, given ]

applying angle sum property on ∆ PQR

|P + |PSR + |PRS = 180°

therefore |P = 180° - (|PSR + |PRS)

=> |P = 180° - (90° + 50°)

=> |P = 180° - 140°

=> |P = 40°

Answer 2

Answer:

Given : angle R = 100° and RS is perpendicular to PQ.

To find : angle P

Now,

In ∆ PQR,

R = 100° => angle PRS = 50° and QRS = 50°

because RS bisects angle R.

In ∆ PRS,

angle P + R + RSP = 180° ( angle sum property of a ∆ )

P + 90° + 59° = 180

P + 140° = 180°

P = 180° - 140°

P = 40°

So, angle P = 40°.