Question

7. In figure 6.44, PQR is an isosceles triangle with PR = QR.
If angle P=80°, find
(i) angle PQR and angle PRQ, and
(ii) the values of angle a and angle b.​

Answer 1

Answer:

Pq=PR(given)

angle q= 80+q=180 (co int angles are supplement )

q=180-80

q=100

value of angle a

180-100 (linear pair)

=100.

Step-by-step explanation:

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Answer 2

Answer:

angles PQR=80°

PRQ=20°

A=100°

B=160°

Step-by-step explanation:

(i)In an  isosceles Δ the angles opposite to equal sides are equal

So ∠QPR=∠PQR = 80°

Sum of angles in a Δ= 180°

= in ΔPQR: ∠RQP+∠PQR +∠PRQ =180°

= 80°+80°+∠PRQ=180°

= ∠PRQ= 180°-160°

= ∠PRQ=20°

(ii) Angles on a line add up to 180° (linear pair)

a: ∠a+∠PQR= 180°

= ∠a= 180-80

∠a=100°

b: ∠b+∠PRQ=180°

= ∠b= 180-20

∠b=160°