Question

1. Given: circle k(O), m

Answer 1

Answer:

1. $m\angle EFO=62^{\circ}\\ \\m\angle EFD=76^{\circ}$

2. $m\angle RUS=65^{\circ}\\ \\m\angle UST=15^{\circ}$

Step-by-step explanation:

Q1. Given circle k(O).

The measure of the arc FE is 56°, this means the measure of the central angle FOE is 56° too.

Consider triangle FOE. This triangle is isosceles triangle with the base FE because FO = EO as radii of the circle.

Angles adjacent to the base of the isosceles triangle are congruent, so

$m\angle EFO=m\angle FEO$

The sum of the measures of all interior angles in the triangle FEO is always 180°, then

$2m\angle EFO+56^{\circ}=180^{\circ}\\ \\2m\angle EFO=124^{\circ}\\ \\m\angle EFO=62^{\circ}$

Angle FDE is inscribed angle subtended on the arc FE, hence its measure is the half of the central angle FOE:

$m\angle FDE=\dfrac{1}{2}\cdot 56^{\circ}=28^{\circ}$

Since FD = ED, thriangle FDE is isosceles triangle with congruent angles adjacent to the base FE. Then

$m\angle EFD=\dfrac{1}{2}(180^{\circ}-28^{\circ})=76^{\circ}$

Q2. If the measure of the arc RU is 50°, then the measure of the central angle ROU is 50° too.

If the measure of the arc UT is 30°, then the measure of the central angle TOU is 30° too.

Triangle ROU is isosceles triangle because RO = UO as radii of the circle. Angles adjacent to the base of the isosceles triangle are congruent, so

$m\angle ORU=m\angle OUR$

The sum of the measures of all interior angles in the triangle ROU is always 180°, then

$2m\angle RUO+50^{\circ}=180^{\circ}\\ \\2m\angle RUO=130^{\circ}\\ \\m\angle RUO=65^{\circ}\\ \\m\angle RUS=m\angle RUO=65^{\circ}$

Angle UST is inscribed angle subtended on the arc UT and has the measure that is half the measure of the central angle TOU:

$m\angle UST=\dfrac{1}{2}m\angle TOU=\dfrac{1}{2}\cdot 30^{\circ}=15^{\circ}$