In triangle ABC, measure of angle B is 20, and that of angle C is 40, length of the bisector of angle A is equal to 2. What is the value of BC - AB?

Question

In triangle ABC, the measure of angle B is 20, and the measure of angle C is 40. Moreover, the length of the bisector of angle A is equal to 2. What is the value of BC - AB?

Answer 1

This is a Math Kangaroo problem for grades 7-8.

Extend side CA past point A until it's equal to CB, call that point B', thus BC=B'C and triangle CBB' is isosceles. angle CB'B equals 80.

Let D be the foot of the angle bisector from vertex A. Find all angles. Notice that triangle ADB and triangle AB'B are congruent by ASA, side AB is shared. Thus, AB' (which is equal to BC-AB) = AD =2.

Answer 2

The way to solve this problem is to find the lengths BC and AB and to then subtract them.

 

First, the angle at A is 120 degrees since the three given angles must equal 180 degrees.

 

Let us label the bisector AP, with P being the termination point on BC.    The bisector then divides triangle ABC into two triangles.    In triangle ABP, the law of sines may be used to solve for AB.   The angle at P is 100 degrees because the angle at A is 60 (half of 120) and the angle at B is 20 (given).   

 

So sin 20/2 = sin 100/AB   => AB = 2*sin 100/sin 20 = 5.759

 

Now we can use the law of sines on triangle ABC to find BC     sin 120/BC = sin 40/AB   => BC = 5.759 * sin 120/sin 40 = 7.759.    BC - AB = 7.759 - 5,759 = 2.000