In an isosceles triangle ABC with base AC = 3 √2, cos C = 1/3 find the height of the SC

Let’s draw the height BH, which is also the median of the triangle, then CH = AH = AC / 2 = 3 * √2 / 2 cm.

In a right-angled triangle, BCH CosC = CH / BC, then BC = CH / CosC = (3 * √2 / 2) / (1/3) = 9 * √2 / 2 cm.

By the Pythagorean theorem, we determine the length of the height BH.

BH ^ 2 = BC ^ 2 – CH ^ 2 = 40.5 – 4.5 = 36.

BH = 6 cm.

Determine the area of the triangle ABC. Savs = AC * VN / 2 = 3 * √2 * 6/2 = 9 * √2 cm2.

Also, the area of the triangle ABC is equal to: Savs = AB * CK / 2, then CK = 2 * Savs / AB = 2 * 9 * √2 / (9 * √2 / 2) = 4 cm.

Answer: The length of the SC height is 4 cm.