Join A and B to O, the centre of the circle
So we have triangle ABO with all 3 sides known. We can use the cosine law to find angle Ø
6^2 = 9^2+9^2-2(9)(9)cosO
162cosØ = 126
cosØ = 126/162 = 7/9
Ø = 38.94...° (I stored it in calculator's memory)
so angleBAO = 70.528779...
angle BAC = 141.057...°
area of triangle ABC
= (1/2)(9)(9)sin 141.057..
= appr 25.5 cm^2
----------------
we could also find it with "exact values"
Let angle OAB = angle OBA = x (isosceles)
then 2x + Ø = 180°
Ø = 180-2x
take cos of both sides:
cosØ = cos(180-2x)
7/9 = cos180cos(-2x) + sin180sin(2x)
7/9 = (-1)(cos(2x) + 0
cos(2x) = -7/9
we know 1 - 2sin^2 x = cos(2x)
1 + 7/9 = 2sin^2 x
sin^2 x = 8/9
sinx = √8/3
cosx = √( 1 - 8/9) = √(/9) = 1/3
sin(2x) = 2sinxcosx = 2(√8/3)(1/3)
area of triangle ABC = (1/2)(9)(9)(2(√8/3)(1/3))
= 9√8 cm^2
or appr 25.5 , the same as above!!!!
area of triangle
In an isosceles triangle ABC in which AB=AC=6cm is inscribed in a circle of radius 9cm .find the area of triangle
1 answer