In the circle shown below, segment BD is diameter and the measure of arc CB is 54°. What is the measure of ∡DBC?

segment BD is a diameter and the measure of arc CB is 54°. Finding the measure of ∡DBC

Solution:

Given, BD is the diameter.

The measure of arc CB is 54°.

The triangle ABC is an isosceles triangle.

segment BD is a diameter and the measure of arc CB is 54°. finding the measure of ∡DBC using angle in a semicircle and the triangle sum property

In an isosceles triangle, two sides and two angles are equal.

From the figure, 

AC = AB → radius of the circle

∠DBC = ∠ACB → angles of the base of the isosceles triangle

∠CAB = 54° → vertex angle of the isosceles triangle

We know, the sum of interior angles = 180°

So, ∠CAB + ∠DBC +∠ACB = 180°

Since, ∠ACB = ∠DBC

54° + 2∠DBC = 180°

2∠DBC = 180°- 54°

∠DBC = 126°/2

∠DBC = 63°

Therefore, the measure of ∠DBC is 63°.

In the circle shown below, segment BD is diameter and the measure of arc CB is 54°. What is the measure of ∡DBC?

Summary:

In the circle shown, segment BD is diameter and the measure of arc CB is 54°. The measure of ∠DBC is 63°.

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