# A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #(pi)/2 #, and the triangle's area is #2 #. What is the area of the triangle's incircle?

The area of the triangle's incircle is

Let

The simplest formula to find the inradius is

where

See this answer of mine for a generalised proof of this. As all triangles are tangential, the formula applies here aswell.

The angle of vertex

Our triangle is visualised below.

As it is a right triangle, its area will be equal to

Using the trigonometric functions, we see that

Using the sum and difference formulas, we find out that

Hence

Then,

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To find the area of the triangle's incircle, we can use the formula (A = rs), where (A) is the area of the triangle, (r) is the radius of the incircle, and (s) is the semi-perimeter of the triangle.

First, we need to find the lengths of the sides of the triangle using the given angles.

Since the angle at vertex (A) is ( \frac{\pi}{12} ) and the angle at vertex (B) is ( \frac{\pi}{2} ), the remaining angle at vertex (C) can be found by subtracting the sum of the other two angles from ( \pi ):

[ \pi - \left( \frac{\pi}{12} + \frac{\pi}{2} \right) = \pi - \left( \frac{\pi}{12} + \frac{6\pi}{12} \right) = \pi - \frac{7\pi}{12} = \frac{5\pi}{12} ]

Now, we can use the Law of Sines to find the lengths of the sides. Let's denote the side opposite vertex (A) as (a), the side opposite vertex (B) as (b), and the side opposite vertex (C) as (c).

[ \frac{a}{\sin\left(\frac{\pi}{12}\right)} = \frac{b}{\sin\left(\frac{\pi}{2}\right)} = \frac{c}{\sin\left(\frac{5\pi}{12}\right)} ]

[ a = \frac{\sin\left(\frac{\pi}{12}\right)}{\sin\left(\frac{\pi}{2}\right)} \times c ] [ b = \frac{\sin\left(\frac{\pi}{2}\right)}{\sin\left(\frac{\pi}{2}\right)} \times c ] [ c = \frac{\sin\left(\frac{5\pi}{12}\right)}{\sin\left(\frac{\pi}{2}\right)} \times c ]

Since ( \sin\left(\frac{\pi}{2}\right) = 1 ), we have: [ a = \sin\left(\frac{\pi}{12}\right) \times c ] [ b = c ] [ c = \sin\left(\frac{5\pi}{12}\right) \times c ]

Now, we can use the given area ((A = 2)) to find the semi-perimeter (s):

[ A = rs ] [ 2 = r \times \left( \frac{a+b+c}{2} \right) ] [ 2 = r \times \left( \frac{\sin\left(\frac{\pi}{12}\right) \times c + c + \sin\left(\frac{5\pi}{12}\right) \times c}{2} \right) ]

Solving for (r) will give us the radius of the incircle, and then we can use the formula for the area of the incircle ((A_{\text{incircle}} = \pi r^2)) to find the area.

This process may involve some algebraic manipulations and trigonometric calculations, but it will yield the area of the triangle's incircle.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- A triangle has corners at #(4 ,3 )#, #(2 ,2 )#, and #(7 ,8 )#. What is the radius of the triangle's inscribed circle?
- A circle has a chord that goes from #( pi)/3 # to #(5 pi) / 12 # radians on the circle. If the area of the circle is #16 pi #, what is the length of the chord?
- A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, vertex B has an angle of #( pi)/6 #, and the triangle's area is #98 #. What is the area of the triangle's incircle?
- A triangle has corners at #(8 ,1 )#, #(4 ,5 )#, and #(6 ,7 )#. What is the area of the triangle's circumscribed circle?
- Points #(2 ,6 )# and #(5 ,3 )# are #(3 pi)/4 # radians apart on a circle. What is the shortest arc length between the points?

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