Given:
Point D is incenter of ΔBCA.
m∠FDG = 136°
To find:
The measure of ∠FHG
Solution:
Angle subtended by the arc at the center is twice that subtended on the remain part of the circle.
[tex]\Rightarrow m\angle FDG = 2 m\angle FHG[/tex]
Divided by 2 on both sides.
[tex]$\Rightarrow \frac{1}{2} m\angle FDG = \frac{1}{2} \times 2 m\angle FHG[/tex]
[tex]$\Rightarrow \frac{1}{2} m\angle FDG = m\angle FHG[/tex]
Substitute m∠FDG = 136°
[tex]$\Rightarrow \frac{1}{2} (136^\circ)= m\angle FHG[/tex]
[tex]$\Rightarrow 68^\circ= m\angle FHG[/tex]
The measure of ∠FHG is 68°.
Answer:
the correct answer is 68
Step-by-step explanation:
you just have to divide 136 by 2 and you will get 68
