# A triangle has vertices A, B, and C. Vertex A has an angle of #pi/2 #, vertex B has an angle of #( pi)/8 #, and the triangle's area is #98 #. What is the area of the triangle's incircle?

Area of Incircle

Next we have to find the hypotenuse and the other side of the right triangle.

Radius of Incircle

#r_i = (-a + b + c ) / 2 = (9 + 21.75 - 23.55) / 2 ~~ 3.6

Area of Incircle

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To find the area of the triangle's incircle, we need to use the formula:

[ \text{Area} = s \cdot r ]

Where ( s ) is the semi-perimeter of the triangle and ( r ) is the radius of the incircle.

First, let's find the lengths of the sides of the triangle. We can use the law of sines since we know two angles and one side:

[ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} ]

Given that angle ( A = \frac{\pi}{2} ), ( B = \frac{\pi}{8} ), and the area is 98, we can find the side lengths.

Since ( A = \frac{\pi}{2} ), ( \sin(A) = 1 ). Also, we know that ( B = \frac{\pi}{8} ), so ( \sin(B) = \sin\left(\frac{\pi}{8}\right) ).

Let's denote the side opposite angle ( A ) as ( a ) and the side opposite angle ( B ) as ( b ):

[ \frac{a}{1} = \frac{b}{\sin\left(\frac{\pi}{8}\right)} ]

[ a = b \cdot \sin\left(\frac{\pi}{8}\right) ]

Now, we can use the area formula for a triangle to find the value of ( b ):

[ \text{Area} = \frac{1}{2} \cdot a \cdot b ]

[ 98 = \frac{1}{2} \cdot b \cdot b \cdot \sin\left(\frac{\pi}{8}\right) ]

[ 196 = b^2 \cdot \sin\left(\frac{\pi}{8}\right) ]

[ b^2 = \frac{196}{\sin\left(\frac{\pi}{8}\right)} ]

[ b = \sqrt{\frac{196}{\sin\left(\frac{\pi}{8}\right)}} ]

Now that we have the value of ( b ), we can find the value of ( a ):

[ a = b \cdot \sin\left(\frac{\pi}{8}\right) ]

Now, we can find the semi-perimeter, ( s ):

[ s = \frac{a + b + c}{2} ]

And finally, we can use the formula ( \text{Area} = s \cdot r ) to find the area of the incircle.

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To find the area of the triangle's incircle, we need to use the formula for the area of a triangle in terms of its inradius ((r)), which is given by (A = rs), where (A) is the area of the triangle and (s) is the semi-perimeter.

First, we need to find the lengths of the sides of the triangle using the given angles. Then we can calculate the semi-perimeter (s) using the formula (s = \frac{a + b + c}{2}), where (a), (b), and (c) are the lengths of the sides.

Given that vertex A has a right angle ((\frac{\pi}{2})), we can use trigonometric ratios to find the lengths of the sides opposite to angles at vertices B and C.

Let's denote:

- (a) as the side opposite angle (\frac{\pi}{8}) at vertex B,
- (b) as the side opposite the right angle at vertex A,
- (c) as the side opposite angle (\frac{\pi}{4}) at vertex C.

From the given information, we can deduce that (b = 98).

Using trigonometric ratios for a right triangle, we can find (a) and (c): [a = b \cdot \tan\left(\frac{\pi}{8}\right)] [c = b \cdot \tan\left(\frac{\pi}{4}\right)]

Once we have (a), (b), and (c), we can calculate (s) and then use the given area ((A = 98)) to solve for the inradius (r).

Finally, the area of the incircle can be found using the formula for the area of a circle ((A_{\text{incircle}} = \pi r^2)), where (r) is the inradius we calculated earlier.

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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.

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