Respuesta :
Answer:
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
Step-by-step 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
Answer:
Step-by-step 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