solve the problem?

Question

an isosceles triangle has two sides of length a, and the angle between them is θ. Write the area A of the triangle in terms of a and θ. Then find the area of the triangle when a = 10 inches and θ = 30°.

please show as much work as possible


thanks for your help

Answer 1

Drop a perp from the apex to the horizontal; now you have two right triangles.
area of a triangle = (1/2) base x height = bh/2
sin (theta/2) = b/a so that b = a sin (theta/2)
cos (theta/2) = h/a so that h = a cos (theta/2)
Area = 2(1/2)[a sin (theta/2)][a cos (theta/2)]
= a^2 sin (t/2) cos (t/2) where t = theta

Answer 2

if you draw the altitude (height) of the triangle to the base,
you can find out the length of the base and height as a trig fn
height = acos(x/2), base = 2asin(x/2), x is the angle in between.
area = 1/2 a cos(x/2) * 2a sin(x/2)
= a^2 cos(x/2)sin(x/2)

If you know the double angle formula, you can reduce it to
area = 1/2 a^2sinx

when a = 10 x = 30
area = 1/2 100 sin(30) = 25