Question

In triangle PQR, if 2∠P=3∠Q=6∠R, find ∠P, ∠Q and ∠R?

Answer 1

2∠P=3∠Q                 3∠Q=6∠R
∠P=3/2∠Q(1)           ∠R=1/2∠Q (2)

By angle sum property in triangle PQR
∠P+∠Q+∠R=180°
3/2∠Q+∠Q+1/2∠Q=180°               (FROM 1 AND 2)
6∠Q/2=180°
3∠Q=180°
∠Q=60°
∠P=3/2×60°=90°        ∠Q=60°
∠R=1/2×60=30°

Answer 2

Answer:

∠P = 90°, ∠Q = 60° and ∠R = 30°.

Step-by-step explanation:

Given:-

In ΔPQR, 2∠P = 3∠Q = 6∠R

To find:-

The values of ∠P, ∠Q and ∠R.

According to the question,

2∠P = 3∠Q = 6∠R

⇒ 2∠P = 3∠Q and 2∠P = 6∠R

⇒ 2/3∠P = ∠Q and 2/6∠P = ∠R

⇒ ∠Q = 2/3∠P and ∠R = 1/3∠P . . . . . (1)

In ΔPQR,

∠P $+$ ∠Q $+$ ∠R = 180° (Angle sum property)

∠P + 2/3∠P + 1/3∠P = 180° (From (1))

                     6/3∠P = 180°

                        2∠P = 180°

                          ∠P = 180/2

                          ∠P = 90°

Substitute the value of ∠P in the equation (1), we get

∠Q = 2/3(90)

      = 2(30)

∠Q = 60°

And

∠R = 1/3(90)

∠R = 30°

Final answer: The values of ∠P = 90°, ∠Q = 60° and ∠R = 30°.

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