Question

what is the measure of ∠abd?
○ 27°
○ 54°
○ 114°
○ 124°
(there is a diagram with triangle dbc, angle at d is (2n)°, angle at c is 60°, and at point b, there is a ray ba with angle (4n + 6)°)

Explanation:

Step1: Use exterior angle theorem

The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. So, \(4n + 6=2n + 60\).

Step2: Solve for n

Subtract \(2n\) from both sides: \(4n-2n + 6=2n-2n + 60\), which simplifies to \(2n+6 = 60\). Then subtract 6 from both sides: \(2n=60 - 6=54\), so \(n = 27\).

Step3: Find \(\angle ABD\)

First, find the measure of the exterior angle \((4n + 6)^{\circ}\), substitute \(n = 27\): \(4\times27+6=108 + 6=114^{\circ}\). Since \(\angle ABD\) and the angle \((4n + 6)^{\circ}\) are supplementary (they form a linear pair), \(\angle ABD=180-(4n + 6)\)? Wait, no. Wait, actually, in the triangle, \(\angle ABD\) is adjacent to the exterior angle? Wait, no, let's re - examine. The exterior angle at B is \((4n + 6)^{\circ}\), and \(\angle ABD\) and the angle at B inside the triangle (let's call it \(\angle DBC\)) and the exterior angle form a linear pair? Wait, no, the triangle has angles at D: \(2n\), at C: \(60^{\circ}\), and at B (inside the triangle) let's call it \(\angle DBC\). Then the exterior angle at B ( \((4n + 6)^{\circ}\)) is equal to \(\angle D+\angle C\) (exterior angle theorem). So \((4n + 6)=2n+60\), we found \(n = 27\), so \(2n=54^{\circ}\) (angle at D). Then, in triangle DBC, the sum of angles is \(180^{\circ}\), so \(\angle DBC=180-(2n + 60)=180-(54 + 60)=66^{\circ}\). Then \(\angle ABD = 180-\angle DBC=180 - 66 = 114^{\circ}\)? Wait, no, wait. Wait, the exterior angle is \((4n + 6)^{\circ}\), and \(\angle ABD\) is adjacent to it? Wait, no, the points are A---B---C on a straight line? Wait, A, B, C are colinear? So \(\angle ABC\) is a straight angle, \(180^{\circ}\). The angle at B inside the triangle is \(\angle DBC\), and \(\angle ABD+\angle DBC = 180^{\circ}\). Also, by exterior angle theorem, \((4n + 6)^{\circ}=\angle D+\angle C=(2n)^{\circ}+60^{\circ}\). We solved \(4n+6 = 2n + 60\) to get \(n = 27\). Then \(\angle D=2n = 54^{\circ}\). Then, in triangle DBC, \(\angle DBC=180-(54 + 60)=66^{\circ}\). Then \(\angle ABD=180 - 66 = 114^{\circ}\).

Answer:

\(114^{\circ}\)