A^2 = b^2 + c^2 - 2bc cos a
= 11^2 + 5^2 - 2*5*11 cos 40
= 7.86 to 2 DP's
to find the remaining angles use the sine rule:-
a / sin A = b / sin B so
7.857/ sin 40 = 11 / sin B
sin B = 11 sin40 / 7.857 = 0.8999
<B = 64 degrees
so <C = 180 - 64-40 = 76 degrees
Answer:
seven hundred and twenty nine
23+8X is the answer to this problem
Answer:
7, 7.1, √51, 7.2
Step-by-step explanation:
√51 is 7.14
The inscribed circle has its center at the point of intersection of the angle bisectors, which also happen to be the medians. Hence the altitude of the triangle is 3 times the radius, or 12 inches.
The side length of this triangle is 2/√3 times the altitude, so the area is
... Area = (1/2)·b·h = (1/2)·(24/√3 in)·(12 in)
... Area = 48√3 in² ≈ 83.1384 in²