An isosceles triangle has two equal sides of lengh 10cm. Let θ be the angle between the two equal sides; Express the area A of the triangle as a function of θ in radians Suppose that θ is increasing at the rate of 10° per minute. How fast is A changing at the instant θ = pi/3? item At what value of θ will the triangle have a maximum area?

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You need to remember the formula of area of triangle such that:

`A(theta) = (10*10*sin theta)/2 =gt A = (100*sin theta)/2`

`A(theta) = 50 sin theta`

You need to differentiate the area function with respect to theta and then you need to find the roots to equation `A'(theta) = 0`  to find the value of theta that maximizes the area of triangle.

`A'(theta) = 50 cos theta`

You need to solve the equation `A'(theta) = 0`  such that:

`50 cos theta = 0 =gt cos theta = 0 =gt theta = pi/2`

You need to find the value of `A'(theta)`  at `theta = pi/3`  such that:

`A'(pi/3) = 50 cos (pi/3) =gt A'(pi/3) = 50*(1/2) =gt A'(pi/3) = 25`

Hence, evaluating how fast A is changing at `theta = pi/3 ` yields `A' (pi/3) = 25 ` and evaluating the angle theta that maximizes the area yields`theta = pi/2` .

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The isosceles triangle has two equal sides of length 10 cm and the angle between the equal sides is equal to `theta` .

It a perpendicular is dropped from the point of intersection of the equal sides to the third side it intersects the side at its center. The area height of the triangle is equal to `10*sin(90 - theta)` = `10*cos theta`

The base of the triangle is equal to `2*10*cos(90 - theta)` = `20*sin theta` . The area of the triangle is equal to `(1/2)*10*cos theta*20*sin theta` = `100*cos theta*sin theta` = `50*sin (2*theta)` . If the equal side is taken as L the area is `(L^2/2)*sin(2*theta)`

As a function of the length of the equal side L and the angle theta between them the area A = `(L^2/2)*sin(2*theta)`

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