Given a triangle ABM, angle A = 30 and angle M = 45, MB = 6cm. find AB.

In triangle ABM, we find the angle B. the sum of all the angles of any triangle is 180 °, so the angle B = 180 ° – A – M = 180 ° – 30 ° – 45 ° = 105 °.

From vertex B of triangle ABM draw height BH to side AM.

Consider a triangle BHM. It is rectangular (since BH is the height) with a right angle BHM. Hence the angle HBM = 180 ° – 90 ° – 45 ° = 45 °. Those. triangle BHM – isosceles.

Angle and side ratios apply in a right-angled triangle. Sine of angle M is the ratio of the opposite leg BH to the hypotenuse BM, i.e. sinM = BH / BM, hence BH = BM * sinM = 6 * sin45 ° = 6 * √ (2) / 2 = 3√ (2).

Consider a triangle ABH. It is also rectangular because BH is the height. Angle A = 30 °. It is known that a leg lying opposite an angle of 30 ° is equal to half the hypotenuse, i.e. BH = 1 / 2AB. Therefore, AB = 2 * BH = 2 * 3√ (2) = 6√ (2).

Answer: AB = 6√ (2).