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Construct  Δ PQR if PQ = 5 cm, m∠ PQR = 105°  and m∠ QRP = 40°.

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Given: m PQR = 105° and m QRP = 40°

Therefore m PQR + m QRP + m QPR = 180°

⇒ 105° + 40° + m QPR = 180°

⇒ 145° + m QPR = 180°

⇒ mQPR = 180° – 145°

⇒ m QPR = 35°

To construct: Δ PQR where m P = 35° , m Q = 105° and PQ = 5 cm.

Steps of construction:

(a) Draw a line segment PQ = 5 cm.

(b) At point P, draw  XPQ = 35° with the help of protractor.

(c) At point Q, draw  YQP = 105° with the help of protractor.

(d) XP and YQ intersect at point R.

It is the required triangle PQR.

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