Question

In the given figure, Triangle PQR and Triangle SQR are isosceles traingles. The equal sides have been shown with similar markings.

a.Find angle PQR and angle PRQ.?
b.Find angle SQR and angle SRQ.?
c.Find the value of x ?​

Answer 1

PQR=PRQ=75°

SQR=SRQ=55°

PQR-SQR=PQS=x

75°-55°=20°

so x=20°

Answer 2

in the figure

angleQ =angle R....(since , sidePQ is equal to sidePR)

let angleQ =angleR =a°

therefore,angleP +angleQ + angle R = 180°......(sum of all angle of triangle os 180°)

30°+a+a=180°

30°+2a=180°

2a=180-30

2a=150°

a=75°

angleQ=angleR=75°

similarly in ∆SQR,

angleSQR=angleSRQ=55°

Now, AngleQ =anglePQS+angleSQR

Therefore , 75°=x°+55°

x=75°-55°=20°

x=20°