Question

48. ABC is a triangle with \(\angle CAB = 15^{\circ}\) and \(\angle ABC = 30^{\circ}\). If M is the midpoint of AB, then \(\angle ACM = ?\)

• A. \(15^{\circ}\)
• B. \(30^{\circ}\)
• C. \(45^{\circ}\)
• D. \(60^{\circ}\)

Hint:

Find angle ACB using the angle sum property, then split it in the ratio of the two base angles since CM goes to the midpoint of AB.

Answer 1

The Correct Option is C

Step 1: Set up the triangle and find angle ACB.
In triangle ABC, \(\angle CAB = 15^{\circ}\) and \(\angle ABC = 30^{\circ}\). Since the angles of a triangle add up to \(180^{\circ}\):
\[ \angle ACB = 180^{\circ} - 15^{\circ} - 30^{\circ} = 135^{\circ} \]

Step 2: Use the fact that M is the midpoint of AB.
Because M is the midpoint of AB, the cevian CM splits the vertex angle \(\angle ACB\) into two parts, \(\angle ACM\) and \(\angle BCM\), in the same ratio as the two base angles of the triangle, \(\angle ABC\) and \(\angle CAB\). This gives:
\[ \angle ACM : \angle BCM = \angle ABC : \angle CAB = 30^{\circ} : 15^{\circ} = 2 : 1 \]

Step 3: Split the 135 degrees in this ratio.
The two parts \(\angle ACM\) and \(\angle BCM\) together make up the full \(135^{\circ}\), and they are in the ratio \(2:1\), with \(\angle ACM\) taking the smaller share, the part lying on the side of the smaller base angle \(\angle CAB\), and \(\angle BCM\) taking the larger share. Splitting \(135^{\circ}\) into 3 equal parts, since \(2+1=3\), gives each part as \(45^{\circ}\). So:
\[ \angle ACM = 1 \times 45^{\circ} = 45^{\circ}, \quad \angle BCM = 2 \times 45^{\circ} = 90^{\circ} \]

Final Answer:
The angle ACM works out to \(45^{\circ}\). \[ \boxed{45^{\circ}} \]