Question

two sides of a triangle have lengths of 12m and 15m. The angle between them is increasing at a rate of 0.06 rad/s. Find the rate at which the area of the triangle is increasing

Answer 1

Explanation:

Area of a triangle = 1/2 × b × h

= 1/2 × b × l × sin theta

Given:

b = 15m

l = 12m

Dtheta/dt = 0.06 rad/s

Area = 1/2 × 15 × 12 × sin theta

= 90 × sin theta

dA/dt = dA/dtheta × dtheta/dt

= 90 cos theta × 0.06

Including pythagoras rule,

Cos pi/3 = 1/2

= 90 × 1/2 × 0.06

= 2.7 m2/s

Answer 2

Incomplete question

Complete question:

Two sides of a triangle have lengths of 12m and 15m. The angle between them is increasing at a rate of 0.06 rad/s. Find the rate at which the area of the triangle is increasing when the angle between the sides of the fixed length is π/4

Answer: 3.81u²/s

Explanation:

Area of a triangle is given as

A = bh/2 . . . eqn1

We have our base to be 12m and our hypotenuse to be 15m and height = ??

Since we don't have h, we would find h using this expression

Sin∅ = opp/hyp = h/15

Therefore, h = 15sin∅. . .eqn 2

But d∅/dt = 0.06rad/s. . .eqn 3

We can now substitute eqn 2 and 3 into eqn 1. We have that

A = bh/2 = 12×15sin∅/2

A = 90sin∅, by differentiating this equation with respect to A, we have

d(A)/dt = 90cos∅d∅/dt

d(A)/dt = 90cos∅(0.06)

∅ = π/4, cosπ/4 = 0.7071

90×0.7071×0.06

= 3.81u²/s