Question

In triangle PQR, angle QPR is acute, PQ = 10 cm and PR = 14 cm.
a) The area of triangle PQR is 48 sq cm. Calculate angle QPR nearest to 1
decimal place.

Answer 1

Given :- In triangle PQR, angle QPR is acute, PQ = 10 cm and PR = 14 cm. The area of triangle PQR is 48 sq cm. Calculate angle QPR nearest to 1 decimal place. ?

Solution :-

Let us assume that, ∠QPR is equal to θ .

we know that,

• Area of a ∆ = (1/2) * a * b * sinθ, where a , b are sides of ∆ and θ is angle between them .

So,

→ Area of ∆PQR = (1/2) * QP * PR * sin(QPR)

→ 48 = (1/2) * 10 * 14 * sin θ

→ 48 = 5 * 14 * sin θ

→ 48 = 70 * sin θ

→ sin θ = (48/70)

→ sin θ = 0.68

→ θ = sin^(-1)(0.68)

→ θ = 43.3° (Ans.)

Learn more :-

In ABC, AD is angle bisector,

angle BAC = 111 and AB+BD=AC find the value of angle ACB=?

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