In triangle ABC the bisector AL is drawn, where L lies on side BC. Angle ALC is 98 degrees and angle ABC is 60 degrees. Find angle ACB in degrees.

Step-by-step solution

1. 1

   Name the unknown you actually want. Set $\angle ACB = x$. Choosing the requested angle as the variable keeps every later equation in terms of $x$ alone, instead of carrying three unknown angles around.

2. 2

   Express $\angle A$ using the angle sum of triangle $ABC$. Since the three angles add to $180^\circ$,

   $\angle A = 180^\circ - 60^\circ - x = 120^\circ - x$

3. 3

   Use the bisector to halve that angle. $AL$ bisects $\angle A$, so the part of it that sits inside triangle $ALC$ is

   $\angle CAL = \frac{120^\circ - x}{2}$

   This is the only place the word bisector is used, and it is what links the big triangle to the small one.

4. 4

   Write the angle sum for triangle $ALC$. Because $L$ lies on segment $BC$, the angle of triangle $ALC$ at $C$ is the same $\angle ACB = x$. So

   $\frac{120^\circ - x}{2} + 98^\circ + x = 180^\circ$

5. 5

   Solve the linear equation.

   $\frac{120 - x}{2} + x = 82 \;\Longrightarrow\; 60 - \frac{x}{2} + x = 82 \;\Longrightarrow\; \frac{x}{2} = 22$

   $x = 44^\circ$

6. 6

   Verify all four angles are consistent. With $x = 44^\circ$ we get $\angle A = 76^\circ$, so each half is $38^\circ$. In triangle $ALC$: $38^\circ + 98^\circ + 44^\circ = 180^\circ$. In triangle $ABC$: $76^\circ + 60^\circ + 44^\circ = 180^\circ$. Both check out.