Part (a)
Answer: 4.9
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Explanation:
Focus on triangle JEA. It is a right triangle with one leg of AE = 9.4 and hypotenuse AJ = 10.6
Use the pythagorean theorem to find the missing leg.
a^2 + b^2 = c^2
9.4^2 + b^2 = 10.6^2
88.36 + b^2 = 112.36
b^2 = 112.36 - 88.36
b^2 = 24
b = sqrt(24)
b = 4.89898
That rounds to roughly 4.9
If your teacher instructs you to round otherwise, then be sure to follow those instructions.
Recall that the incenter is the center of the incircle. This special circle is the largest possible that is completely contained in the triangle. No parts of the circle spill outside the triangle. If you had a triangular box of this size and shape, then figuring out the incircle will tell you the largest possible circle you can pack inside.
Because J is the center of the incircle, this means the radii JE and JF are the same length. We just computed that JE = 4.9 roughly, so JF = 4.9 as well.
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Part (b)
Answer: 27.6
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Explanation:
The process of finding the incenter involves constructing the angle bisectors.
- Segment JB bisects angle ABC
- Segment JC bisects angle BCA
- Segment JA bisects angle CAB
Because angle ABC is split in half, and because the diagram shows angle JBC = 33.9, we know that angle JBA is also 33.9 degrees.
Overall, angle ABC = angle JBC+angle JBA = 33.9 + 33.9 = 67.8 degrees.
Through similar logic, you should find that angle BCA is 57 degrees.
The missing angle CAB is 180-67.8-57 = 55.2 degrees
Then cut that in half to get the answer we want
55.2/2 = 27.6
