In the diagram of circle r, m∠FGH is 50°. What is m∠FEH?

Solution:

Given, m∠FGH = 50°

We have to find the value of m∠FEH.

From the figure, 

∠FGH is the exterior angle.

We know that the measurement of the exterior angle is the semi-difference of the arcs which comprises.

So, ∠FGH = \(\frac{1}{2}(arcFEH-arcFH)\)

\(50^{\circ}=\frac{1}{2}(arcFEH-arcFH)\)

\(100^{\circ}=(arcFEH-arcFH)\) --- (1)

We know, \(arcFEH+arcFH=360^{\circ}\)

\(arcFH=360^{\circ}-arcFEH\) --- (2)

Substituting equation (2) in (1)

\(100^{\circ}=(arcFEH-(360^{\circ}-arcFEH))\)

\(100^{\circ}=arcFEH-360^{\circ}+arcFEH\)

\(100^{\circ}=2arcFEH-360^{\circ}\)

\(2arcFEH=360^{\circ}+100^{\circ}\)

\(2arcFEH=460^{\circ}\)

\(arcFEH=\frac{460^{\circ}}{2}\)

arcFEH = 230°

Therefore, the m∠FEH is 230°.

In the diagram of circle r, m∠FGH is 50°. What is m∠FEH?

Summary:

In the diagram of circle r, m∠FGH is 50°, then m∠FEH is 230°.