Question

NO LINKS!! URGENT HELP PLEASE!! ​

Answer 1

`\text{a.} \quad m\angle NLM=93^(\circ)`

`\text{c.} \quad m\angle FHG=31^(\circ)`

Explanation:

The inscribed angle in the given circle is ∠NLM.

The intercepted arc in the given circle is arc NM = 186°.

According to the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of the intercepted arc.

Therefore:

`m\angle NLM=(1)/(2)\overset{\frown}{NM}`

`m\angle NLM=(1)/(2) \cdot 186^(\circ)`

`\boxed{m\angle NLM=93^(\circ)}`

`\hrulefill`

According to the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of the intercepted arc. Therefore:

`m\angle HFG=(1)/(2)\overset{\frown}{HG}`

`m\angle HFG=(1)/(2)\cdot 118^(\circ)`

`m\angle HFG=59^(\circ)`

As line segment FH passes through the center of the circle, FH is the diameter of the circle. Since the angle at the circumference in a semicircle is a right angle, then:

`m\angle FGH = 90^(\circ)`

The interior angles of a triangle sum to 180°. Therefore:

`m\angle FHG + m\angle HFG + m\angle FGH =180^(\circ)`

`m\angle FHG + 59^(\circ) + 90^(\circ) =180^(\circ)`

`m\angle FHG +149^(\circ) =180^(\circ)`

`\boxed{m\angle FHG =31^(\circ)}`