In a triangle, the length of side AB is 2, the angle A is 60 degrees, and the angle B is 70 degrees.

In a triangle, the length of side AB is 2, the angle A is 60 degrees, and the angle B is 70 degrees. Point D is taken on the AC side, so AD = 1. Find the degree measure of the DBC angle.

Given:

AB = 2

AD = 1

A = 600

B = 700

Find: Angle DBC

Decision:

1) Consider a triangle ABD. By the cosine theorem, we find the side of BD:

BD ^ 2 = AB ^ 2 + AD ^ 2 – 2AB * AD cos A

BD2 = 4 + 1 – 2 * 2 * 1 * ½ = 3

BD = √3

2) By the cosine theorem, we find the angle ABD in the triangle ABD:

AD ^ 2 = AB ^ 2 + B ^ D2 – 2 AB * BD cos ABD

cos ABD = (AB ^ 2 + BD ^ 2 – AD ^ 2) / 2 AB * BD

cos ABD = (4 + 3 – 12) / 2 * 2 * √3 = 6 / 4√3 = √3 / 2

Hence: angle ABD = 300

3) ABC angle = ABD angle + DBC angle

angle DBC = angle ABC – angle ABD = 700 – 300 = 400

Answer: 400.