Answer :
Certainly! Let's solve this step-by-step.
1. Understanding the given data:
- We have an isosceles triangle [tex]\( \triangle ABC \)[/tex] inscribed in a circle.
- The base of the triangle is [tex]\( \overline{BC} \)[/tex].
- The measure of the arc [tex]\( \overparen{BC} \)[/tex] is [tex]\( 102^\circ \)[/tex].
2. Determining the central angle:
- The angle subtended by the arc [tex]\( \overparen{BC} \)[/tex] at the center of the circle is equal to the measure of the arc itself.
- Therefore, the central angle [tex]\( \angle BOC \)[/tex] is [tex]\( 102^\circ \)[/tex].
3. Determining the angle at the circumference:
- The angle subtended by the arc [tex]\( \overparen{BC} \)[/tex] at the circumference of the circle is half the measure of the central angle.
- Thus, [tex]\( \angle BAC \)[/tex] is [tex]\( \frac{102^\circ}{2} = 51^\circ \)[/tex].
4. Identifying the remaining base angles:
- Since [tex]\( \triangle ABC \)[/tex] is isosceles with base [tex]\( \overline{BC} \)[/tex], the angles at vertices [tex]\( B \)[/tex] and [tex]\( C \)[/tex] (the base angles) are equal.
- Let [tex]\( \angle ABC = \angle ACB = x \)[/tex].
5. Using the triangle angle sum property:
- The sum of the angles in any triangle is [tex]\( 180^\circ \)[/tex].
- We know that [tex]\( \angle BAC + \angle ABC + \angle ACB = 180^\circ \)[/tex].
- Substituting the known values, we get:
[tex]\[ 51^\circ + x + x = 180^\circ \][/tex]
[tex]\[ 51^\circ + 2x = 180^\circ \][/tex]
6. Solving for the base angles:
- Subtract [tex]\( 51^\circ \)[/tex] from both sides:
[tex]\[ 2x = 180^\circ - 51^\circ \][/tex]
[tex]\[ 2x = 129^\circ \][/tex]
- Divide by 2:
[tex]\[ x = \frac{129^\circ}{2} = 64.5^\circ \][/tex]
Therefore, the measures of the angles in [tex]\( \triangle ABC \)[/tex] are:
- [tex]\( \angle BAC = 51^\circ \)[/tex]
- [tex]\( \angle ABC = 64.5^\circ \)[/tex]
- [tex]\( \angle ACB = 64.5^\circ \)[/tex]
Thus, the angles of the triangle [tex]\( \triangle ABC \)[/tex] are [tex]\( 51^\circ \)[/tex], [tex]\( 64.5^\circ \)[/tex], and [tex]\( 64.5^\circ \)[/tex].
1. Understanding the given data:
- We have an isosceles triangle [tex]\( \triangle ABC \)[/tex] inscribed in a circle.
- The base of the triangle is [tex]\( \overline{BC} \)[/tex].
- The measure of the arc [tex]\( \overparen{BC} \)[/tex] is [tex]\( 102^\circ \)[/tex].
2. Determining the central angle:
- The angle subtended by the arc [tex]\( \overparen{BC} \)[/tex] at the center of the circle is equal to the measure of the arc itself.
- Therefore, the central angle [tex]\( \angle BOC \)[/tex] is [tex]\( 102^\circ \)[/tex].
3. Determining the angle at the circumference:
- The angle subtended by the arc [tex]\( \overparen{BC} \)[/tex] at the circumference of the circle is half the measure of the central angle.
- Thus, [tex]\( \angle BAC \)[/tex] is [tex]\( \frac{102^\circ}{2} = 51^\circ \)[/tex].
4. Identifying the remaining base angles:
- Since [tex]\( \triangle ABC \)[/tex] is isosceles with base [tex]\( \overline{BC} \)[/tex], the angles at vertices [tex]\( B \)[/tex] and [tex]\( C \)[/tex] (the base angles) are equal.
- Let [tex]\( \angle ABC = \angle ACB = x \)[/tex].
5. Using the triangle angle sum property:
- The sum of the angles in any triangle is [tex]\( 180^\circ \)[/tex].
- We know that [tex]\( \angle BAC + \angle ABC + \angle ACB = 180^\circ \)[/tex].
- Substituting the known values, we get:
[tex]\[ 51^\circ + x + x = 180^\circ \][/tex]
[tex]\[ 51^\circ + 2x = 180^\circ \][/tex]
6. Solving for the base angles:
- Subtract [tex]\( 51^\circ \)[/tex] from both sides:
[tex]\[ 2x = 180^\circ - 51^\circ \][/tex]
[tex]\[ 2x = 129^\circ \][/tex]
- Divide by 2:
[tex]\[ x = \frac{129^\circ}{2} = 64.5^\circ \][/tex]
Therefore, the measures of the angles in [tex]\( \triangle ABC \)[/tex] are:
- [tex]\( \angle BAC = 51^\circ \)[/tex]
- [tex]\( \angle ABC = 64.5^\circ \)[/tex]
- [tex]\( \angle ACB = 64.5^\circ \)[/tex]
Thus, the angles of the triangle [tex]\( \triangle ABC \)[/tex] are [tex]\( 51^\circ \)[/tex], [tex]\( 64.5^\circ \)[/tex], and [tex]\( 64.5^\circ \)[/tex].
