Question

In triangle PQR,PQ=PR and Ext R =130. Then find Ext P +Ext Q

Answer 1

Answer:

By using the theorem,If two sides of a triangle are equal then the opposite angles to the sides are equal.

⇒ If PQ=PR then ∠Q=∠R

In △PQR,

⇒∠P+∠Q+∠R=180

∘

⇒∠P+∠Q+∠R=180

∘

since ∠Q=∠R=65

∘

(given)

⇒∠P+2×65

∘

=180

∘

⇒∠P=180

∘

−130

∘

=50

∘

∴∠P=50

∘

Step-by-step explanation:

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Answer 2

❥ Question :

In triangle PQR,PQ=PR and Ext R =130. Then find Ext P +Ext Q

❥ Given :

PQ=PR and Ext R =130

❥ To Find :

Ext P +Ext Q

❥ Answer :

Ext P +Ext Q =  230

❥ Calculation :

⇒ If PQ=PR then ∠Q =∠R

⇒∠P+∠Q+∠R = 180°

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If ∠Q =∠R , then ∠Q+∠R = 2∠R (even you can take 2∠Q)

⇒∠P+2∠R = 180°

Ext R =130°, so :

∠P+∠Q =130°

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as ∠Q =∠R ,

∠P+∠Q =130°

(or)

∠P+∠R =130° are same

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∠P+2∠R = ∠P+∠R+∠R

130° + ∠R = 180°

∠R = 180°-130°

∠R = 50°

as ∠Q =∠R,

∠Q = 50°

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∠P+2∠R = 180°

∠P+2×50° = 180°

∠P+100° = 180°

∠P= 180° - 100°

∠P= 80°

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$\huge\black \boxed{\bf{\angle P= 80^\circ,\angle Q = 50^\circ,\angle R = 50^\circ}}$

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Ext R = ∠P + ∠Q

         = 80° + 50°

         = 130°

Ext P = ∠R + ∠Q

         = 50° + 50°

         = 100°

Ext Q = ∠P + ∠R

          = 80° + 50°

          = 130°

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$\large\black \boxed{\bf{Ext\:\:R=130,Ext\:\:P=100,Ext\:\:Q=130}}$

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Ext P +Ext Q = 100 + 130

                     = 230

$\large\black \boxed{\bf{Ext\:\: P +Ext\:\: Q = 230}}$

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