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`        Find the magnitude of vertex angle of isosceles triangle of given area A such that radius r of circle inscribed into the triangle is maximum`
5 years ago Sumit Majumdar
IIT Delhi
137 Points
```							Dear student,
Here is help with a solution to finding angles of a triangle having

maximum area if it is drawn within a circle so that the triangle's

vertices just touch the circle's perimeter (i.e., a triangle inscribed

within a circle).

First draw an isosceles triangle within a circle having an arbitrary

. The angle at the top of the triangle is assigned to Phi. One half of

this angle is Phi/2. The two equal angles at the base of the triangle

are assigned to Theta. The two equal side lengths of this triangle are

assigned to L. The base of the triangle is assigned to B. The height

of the triangle is assigned to H.

Now, it is well known that the Area (A) of any triangle is:

A = (1/2)BH.

To find the angles associated with the inscribed triangle (described

above) having a maximum area, one may first re-express the area

formula above in terms involving only Phi (or Phi/2) and R.

You should be able to show that:

L/2 = Rcos(Phi/2), so

L = 2Rcos(Phi/2), and

B = 2[2Rcos(Phi/2)sin(Phi/2)] = 4Rcos(Phi/2)sin(Phi/2), and

H = 2Rcos(Phi/2)cos(Phi/2) = 2Rcos^2(Phi/2).

Therefore:

A = (1/2)BH = (1/2)[4Rcos(Phi/2)sin(Phi/2)][2Rcos^2(Phi/2)], or

A = 4R^2[cos^3(Phi/2)sin(Phi/2)] .

Now, one may take the first derivative of A with respect to Phi [i.e.,

dA/dPhi] by holding R constant because the angles associated with the

inscribed triangle having a maximum area are independent of the

circle's size. If [dA/dPhi] is set to equal zero, one may solve for

the value of Phi that satisfies this condition. Once that value of Phi

is known, one may solve for Theta because: Phi + 2Theta = 180. Then,

all the angles associated with the triangle having a maximum area will

be found.

Upon simplification, one should find that:

[dA/dPhi] = 2R^2cos^2(Phi/2)[cos^2(Phi/2) - 3sin^2(Phi/2)].

Setting: [cos^2(Phi/2) - 3sin^2(Phi/2)] = 0, one can show that:

Phi/2 = 30 deg, so:

Phi = 60 deg, therefore,

2Theta = 120 deg, and,

Theta = 60 deg.

Therefore, the inscribed triangle having a maximum area is an

equilateral triangle

Regards

Sumit

```
5 years ago
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