In triangle ABC the bisector AL is drawn, angle ALC is 98 degrees and angle ABC is 60 degrees. Find angle ACB in degrees.

Step-by-step solution

1. 1

   Introduce the unknown and express angle $A$ through it. Let

   $\angle ACB=x$

   The angle sum of triangle $ABC$ then forces

   $\angle BAC=180^{\circ}-60^{\circ}-x=120^{\circ}-x$

   Working with one unknown rather than two is what makes this a single linear equation at the end.

2. 2

   Use the bisector to halve angle $A$. By definition $AL$ splits $\angle BAC$ into two equal parts, so

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

3. 3

   Write the angle sum for the small triangle $ALC$. Because $L$ lies on segment $BC$, the angle of triangle $ALC$ at $C$ is the same as $\angle ACB$, namely $x$ — that identification is the step that ties the two triangles together. Hence

   $\angle CAL+\angle ALC+\angle ACL=180^{\circ}\ \Longrightarrow\ \frac{120^{\circ}-x}{2}+98^{\circ}+x=180^{\circ}$

4. 4

   Solve the linear equation. Move the $98^{\circ}$ across and expand the fraction:

   $\frac{120-x}{2}+x=82\ \Longrightarrow\ 60-\frac{x}{2}+x=82\ \Longrightarrow\ 60+\frac{x}{2}=82\ \Longrightarrow\ \frac{x}{2}=22\ \Longrightarrow\ x=44$

5. 5

   Check the whole configuration. With $x=44^{\circ}$: $\angle BAC=120^{\circ}-44^{\circ}=76^{\circ}$, so $\angle CAL=\angle BAL=38^{\circ}$. Triangle $ALC$ closes as $38^{\circ}+98^{\circ}+44^{\circ}=180^{\circ}$. Triangle $ABL$ has $\angle ALB=180^{\circ}-98^{\circ}=82^{\circ}$ and closes as $60^{\circ}+38^{\circ}+82^{\circ}=180^{\circ}$. Triangle $ABC$ closes as $60^{\circ}+76^{\circ}+44^{\circ}=180^{\circ}$. All three angle sums are exact, so $\angle ACB=44^{\circ}$.