Question

In the isosceles ∆PQR , ∠P and ∠Q are equal. ∠PRS is an exterior angle of ∆PQR. The
measures of ∠PQR and ∠PRS are (x – 10)0 and (2x + 40)0 respectively. Find the measures of
∠PRQ and ∠PRS. Also find measures of ∠P and ∠Q​

Answer 1

Step-by-step explanation:

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Question 1:

In the given figure, ∠∠ACD is an exterior angle of Δ∆ABC. ∠∠B = 40°, ∠∠A = 70°.



Find the measure of ∠∠ACD.

 

ANSWER:

In Δ∆ABC,

∠ACD = ∠A + ∠B   (Exterior angle property)

= 70∘ + 40∘          

= 110∘

Hence, the measure of  ∠∠ACD is 110

Answer 2

Given:

An isosceles triangle PQR.

∠P = ∠Q

Exterior angle = ∠PRS

Measure of ∠PQR = $x-10$

Measure of ∠PRS = $2x+40$

To find:

Measures of ∠PRQ, ∠PRS, ∠P and ∠Q.

Solution:

In an isosceles triangle, two angles and their opposite sides are equal.

In ΔPQR, ∠P and ∠Q are equal. It is given that the measure of

∠PQR =∠Q = $x-10$

Hence, measure of ∠P = $x-10$

Now, ∠PRQ and ∠PRS lie on the same line. Hence, their angle sum is supplementary.

$\angle PRQ+\angle PRS=180$

$\angle PRQ+2x+40=180$

$\angle PRQ=180-40-2x$

$\angle PRQ=140-2x$

Measures of ∠P and ∠Q are equal and they are $(x-10)^{0}$. Measure of $\angle PRQ=(140-2x)^{0}$ and measure of ∠PRS = $(2x+40)^{0}$