Question

Given circle O with diameter AC, tangent AD, and the measure of arc BC is 74 degrees, find the measures of all other indicated angles.

Answer 1

We want to find the measure of the angles 1 to 8, given that the diameter is AC and the measure of the Arc BC is 74°.

The angle 5, ∡BOC is central and it is equal to the measure of the arc it intercepts, the arc BC. Thus the angle 5 is 74°.

The angle 4, ∡AOB also is central, and it is equal to the measure of the arc AB. As the line AC is the diameter of the circle O, the arc AC is equal to 180°, and thus, the sum of the angles 4 and 5 will be 180°:

`\begin{gathered} \measuredangle4+\measuredangle5=180^(\circ) \\ \measuredangle4=180^(\circ)-\measuredangle5=180^(\circ)-74^(\circ)=106^(\circ) \end{gathered}`

Thus, the angle 4 is 106°.

The angle 6 is an inscribed angle, and thus it is half of the arc it intersects, the arc AB. This means that the angle 6 is 106°/2=54°.

The angle 2 also is an inscribed angle, half of the arc BC, and thus, the angle 2 is 74°/2=37°.

Now, the triangle BOC has the angles 5, 6 and 7, and the sum of those angles is 180°. This means that:

`\begin{gathered} \measuredangle5+\measuredangle6+\measuredangle7=180^(\circ) \\ 74^(\circ)+54^(\circ)+\measuredangle7=180^(\circ) \\ 128^(\circ)+\measuredangle7=180^(\circ) \\ \measuredangle7=180^(\circ)-128^(\circ)=52^(\circ) \end{gathered}`

Thus, the angle 7 is 52°.

Following a same argument, we can get the angle 8, as being part of the triangle AOB.

`\begin{gathered} \measuredangle2+\measuredangle4+\measuredangle8=180^(\circ) \\ \measuredangle8=180^(\circ)-37^(\circ)-106^(\circ)=37^(\circ) \end{gathered}`

This means that the angle 8 is 37°.

As the line AD is tangent to the circle O, this means that the lines AC and AD are perpendicular, and thus, the angle 1 is 90°.

Lastly, as the angles 1, 2 and 3 are coplanar, their sum is 180°. This is:

`\begin{gathered} \measuredangle1+\measuredangle2+\measuredangle3=180^(\circ) \\ \measuredangle3=180^(\circ)-\measuredangle1-\measuredangle2 \\ \measuredangle3=180^(\circ)-90^(\circ)-37^(\circ)=180^(\circ)-127^(\circ)=53^(\circ) \end{gathered}`

Thus, the angle 3 is 53°.