Question

An isosceles triangle has an area of 24 yd², and the angle between the two equal sides is 166°. Find the length of the two equal sides.

Answer 1

Let x = base of triangle
Let y = height of triangle

Area of triangle is (1/2) base times height, so
A = (1/2)xy
24 = (1/2)xy
48 = xy : eqn(1)

If the angle between the two equal sides is 166°, then the remaining angles are:
1/2(180° - 166°) = 7° each

In your drawing of triangle,
tan(angle) = opposite / adjacent
tan(7°) = height of triangle / 1/2 of base of triangle
tan(7°) = y / (1/2x) : eqn(2)

Now you have two equations, two unknowns.
From eqn(2), we can rewrite this as:
tan(7°) = y / (1/2 x)
y = 0.1228 (1/2 x)
y = 0.06138x

Substitute this to eqn(1):
48 = xy
48 = x(0.06138x)
48 = 0.06138x^2
x^2 = 48 / 0.06138
x^2 = 781.857
x = 27.96

Thus, the height is
48 = (27.96)y
y = 1.72

And the length of an equal side is (use Pythagorean Theorem):
(length of an equal side)^2 = (1/2 x)^2 + y^2
(length of an equal side)^2 = (0.5 * 27.96)^2 + 1.72^2

Continue solving for the length, units in yards. Hope this helps~ `u`