In an isosceles triangle ABC (AB = BC), the base is 12 cm, and the height drawn to the base is 8 cm.

In an isosceles triangle ABC (AB = BC), the base is 12 cm, and the height drawn to the base is 8 cm. Find the sine, cosine and tangent of the angles at the base.

The height of the BM divides the base in half in an isosceles triangle. It turns out AM = CM = AC / 2 = 12/2 = 6 cm. We have two right-angled triangles ABM and BMC with two known sides – legs and an unknown – hypotenuse. Let’s find it using the Pythagorean theorem:
AB^2 = BC^2 = BM^2 + AM^2 = 6 * 6 + 8 * 8 = 100, then AB = BC = 10 cm
Now we find the sine, cosine and tangent, respectively using the following formulas:
Sin A = BM / AB = 8/10 = 0.8
Cos A = AM / AB = 6/10 = 0.6
Tg A = Sin A / Cos A = 0.8 / 0.6 = 1 1/3