In an isosceles triangle ABC, the base of BC is 18cm, the medians BN and CM intersect at point O, and the angle

| September 8, 2021 | education

In an isosceles triangle ABC, the base of BC is 18cm, the medians BN and CM intersect at point O, and the angle OBC = 30 degrees. Find the medians. Find through the cosine theorem.

In an isosceles triangle, the medians drawn to the lateral side are equal, CM = BN. Since the medians at point O are divided into segments in the ratio of 2/1, then OB = OС, and the ВOС triangle is isosceles, and the angle ВCO = OBC = 30, then the angle ВOС = (180 – 30 – 30) 120.

We apply the cosine theorem to the BOC triangle.

BC ^ 2 = OB ^ 2 + OS ^ 2 – 2 * OB * OS * Cos120.

Since OB = OС, then:

ВС ^ 2 = 2 * ОВ ^ 2 – 2 * ОВ ^ 2 * Cos120.

324 = 2 * OB ^ 2 * (1 – (-1/2)).

324 = 2 * OB ^ 2 * 3/2.

ОВ ^ 2 = 324/3 = 108 cm.

ОВ = 6 * √3 cm.

Since OB / ON = 2/1, then ON = OB / 2 = 6 * √3 / 2 = 3 * √3 cm.

ВN = ОВ + ON = 6 * √3 + 3 * √3 = 9 * √3 cm.

Answer: The length of the medians is 9 * √3 cm.



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