Expert: Sherman D. - 3/28/2005

Question
1) find the measure, to the nearest tenth of a degree, of the vertex angle of an isosceles triangle with a base of length 8 and legs of length 5.

2) the area of triangle ABC is 45 sq. units. if a=10 and b=15, find the measure of angle C to the nearest degree.

Answer
1.)
on this, you are saying
5,5,8

a^2 = b^2 + c^2 - 2bc(cosA)
8^2 = 5^2 + 5^2 - 2(5 * 5)(cosA)
64 = 25 + 25 - 2(25)cos(A)
64 = 50 - 50cos(A)
14 = -50cos(A)
(-14/50) = cos(A)
A = 106.2602047

with this being the largest side therefore the largest angle, to find the next 2 angles, use this

2x + 106.2602047 = 180
2x = 73.73979529
x = 36.86989765

the answer is

36.9°, 36.9°, 106.3°

To prove i was right about the other 2 angles.

5^2 = 8^2 + 5^2 - 2(4 * 5)cos(B)
25 = 64 + 25 - 2(40)cos(B)
-64 = -80cos(B)
(4/5) = cos(B)
B = 36.86989765

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2.)
Can you tell me what kind of triangle this is. If it is a right triangle, i get
45 = (1/2)ab
45 = (1/2)(10)(15)
45 = (1/2)(150)
45 = 75
Which doesn't work

If it is a isosceles, i get 50 one way and 71 another way.

If this is scalene, then i am having trouble getting the other side to complete this problem.