Question

Two sides of a triangle are 30 cm. and 40 cm. respectively. How fast is the area of the triangle increasing if the angle between the sides is 60 degrees and is increasing at the rate of 4 degrees/sec?

Answer 1

Answer:

Let ABC be the given triangle.

Let AB=c=4 cm and

AC=b=5 cm and θ be the angle between AB and AC at time t.

Given,

dt

dθ

=0.06 rad/sec.

Area of the triangle =

2

1

bc sinθ=

2

1

×4×5sinθ

⇒A=10sinθ

Differenting w.r.t.

′

θ

′

we get,

dt

dA

=10cosθ

dt

dθ

When θ=

3

π

,

dt

dA

=10cos

3

π

(0.06)

=10(

2

1

)(0.06)=0.3 m

2

/sec.

∴ the rate of increase in area =0.3 m

2

/sec.

Answer 2

Answer:

The triangle is increasing at the rate of $0.24 m^{2}/sec$

CONCEPT

Area of a triangle = one side of triangle × another side of triangle × sin(angle between the sides used in finding the area of triangle)

GIVEN

• One side of triangle = 30cm
• Another side of triangle =40cm
• Angle between the sides = 60 degree
• Rate in change of angle = 4 degrees/sec

TO FIND

Rate in change of area of triangle

SOLUTION

Area of triangle(A) = $30.40.$sin(∅)

A=1200sin(∅)

differentiating the above equation with respect to t(time)

$\frac{dA}{dt}=1200$cos∅$\frac{d}{dt}$∅

$\frac{dA}{dt}=1200 cos 60$$\frac{d}{dt}$∅

$\frac{dA}{dt}=1200.\frac{1}{2}.\frac{d}{dt}$∅

$\frac{dA}{dt}=1200.\frac{1}{2}.4$

$\frac{dA}{dt}=2400$ $cm^{2} /sec$

$\frac{dA}{dt}=0.24 m^{2}/sec$

Hence the rate of increasing of triangle is $0.24 m^{2}/sec$.

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Thank you