16. In the given figure, AB=AC and D is a point on BC such that angle BAD = 30 degrees and angle ADC = 110 degrees. Find angle BAC.

Solution:

Let angle ABC = angle ACB = x.

In triangle ABD, angle ADB = 180 - angle ADC = 180 - 110 = 70 degrees.

In triangle ABD, angle BAD + angle ABD + angle ADB = 180 degrees.

30 + x + 70 = 180

x + 100 = 180

x = 180 - 100 = 80 degrees.

So, angle ABC = angle ACB = 80 degrees.

In triangle ABC, angle BAC + angle ABC + angle ACB = 180 degrees.

angle BAC + 80 + 80 = 180

angle BAC + 160 = 180

angle BAC = 180 - 160 = 20 degrees.

Answer: Angle BAC = 20 degrees.