Question

find ∠bac and ∠acb in the figure shown alongside. 60° + 50°+90°≠180° so this is not possible

Explanation:

Step1: Recall angle - sum property of a triangle

The sum of interior angles of a triangle is 180°. In \(\triangle ABC\), we know one exterior angle \(= 130^{\circ}\). The interior - angle adjacent to this exterior angle is \(180 - 130=50^{\circ}\) (linear - pair of angles).

Step2: Calculate \(\angle BAC\)

We know \(\angle ABC = 60^{\circ}\), and the third angle of the triangle (adjacent to the 130° exterior angle) is 50°. Let \(\angle BAC=x\). Using the angle - sum property of a triangle \(\angle ABC+\angle BAC+\angle ACB = 180^{\circ}\). We already found one of the interior angles of the triangle adjacent to the 130° exterior angle is 50° and \(\angle ABC = 60^{\circ}\). So \(x+60^{\circ}+50^{\circ}=180^{\circ}\), then \(x=\angle BAC=180-(60 + 50)=70^{\circ}\).

Step3: Calculate \(\angle ACB\)

We found that the angle adjacent to the 130° exterior angle (let's call it \(\angle ACB\)) is 50° (since they form a linear pair, \(180 - 130 = 50^{\circ}\)).

Answer:

\(\angle BAC = 70^{\circ}\), \(\angle ACB = 50^{\circ}\)