A triangle has vertices A, B, and C. Vertex A has an angle of #pi/12 #, vertex B has an angle of #pi/8 #, and the triangle's area is #5 #. What is the area of the triangle's incircle?

Answer 1

The area of the incircle is #=1.21u^2#

The area of the triangle is #A=5#

The angle #hatA=1/12pi#

The angle #hatB=1/8pi#

The angle #hatC=pi-(1/12pi+1/8pi)=19/24pi#

The sine rule is

#a/(sin hat (A))=b/sin hat (B)=c/sin hat (C)=k#

So,

#a=ksin hatA#

#b=ksin hatB#

#c=ksin hatC#

Let the height of the triangle be #=h# from the vertex #A# to the opposite side #BC#

The area of the triangle is

#A=1/2a*h#

But,

#h=csin hatB#

So,

#A=1/2ksin hatA*csin hatB=1/2ksin hatA*ksin hatC*sin hatB#

#A=1/2k^2*sin hatA*sin hatB*sin hatC#

#k^2=(2A)/(sin hatA*sin hatB*sin hatC)#

#k=sqrt((2A)/(sin hatA*sin hatB*sin hatC))#

#=sqrt(10/(sin(1/12pi)*sin(1/8pi)*sin(19/24pi)))#

#=12.88#

Therefore,

#a=12.88sin(1/12pi)=3.33#

#b=12.88sin(1/8pi)=4.93#

#c=12.88sin(19/12pi)=7.84#

The radius of the incircle is #=r#

#1/2*r*(a+b+c)=A#

#r=(2A)/(a+b+c)#

#=10/(16.1)=0.62#

The area of the incircle is

#area=pi*r^2=pi*0.62^2=1.21u^2#

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Answer 2

To find the area of the triangle's incircle, you can use the formula:

[ \text{Area} = \text{Semiperimeter} \times \text{Inradius} ]

where the semiperimeter is given by:

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

and ( a ), ( b ), and ( c ) are the lengths of the triangle's sides.

To find ( a ), ( b ), and ( c ), you can use the Law of Sines:

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

Given that the angles are ( \frac{\pi}{12} ), ( \frac{\pi}{8} ), and ( \frac{\pi}{3} ) (since the sum of angles in a triangle is ( \pi )), and the area is ( 5 ), you can solve for the sides of the triangle.

Once you have the sides of the triangle, you can calculate the semiperimeter. Then, you can use the formula for the area of the incircle to find the inradius. Finally, plug the semiperimeter and inradius into the formula for the area of the incircle to find the area.

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Answer from HIX Tutor

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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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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