Answer:
[tex] \huge \orange{m\widehat {FGH} =238\degree} [/tex]
Step-by-step explanation:
- Quadrilateral AFGH is inscribed in a circle.
- So, it is a cyclic quadrilateral.
- Opposite angles of a cyclic quadrilateral are supplementary.
[tex] \therefore m\angle FGH + m\angle FAH = 180\degree [/tex]
[tex] \therefore (21x-2)\degree + (38x +5)\degree = 180\degree [/tex]
[tex] \therefore (21x-2+38x +5)\degree = 180\degree [/tex]
[tex] \therefore 59x+3= 180[/tex]
[tex] \therefore 59x= 180-3 [/tex]
[tex] \therefore 59x= 177 [/tex]
[tex] \therefore x= \frac{177}{59}[/tex]
[tex] \therefore x=3[/tex]
Now, by inscribed angle theorem:
[tex] m\angle FAH = \frac{1}{2} m\widehat {FGH} [/tex]
[tex] (38x +5)\degree = \frac{1}{2} m\widehat {FGH} [/tex]
[tex] 2(38\times 3+5)\degree = m\widehat {FGH} [/tex]
[tex] 2(119)\degree = m\widehat {FGH} [/tex]
[tex] 238\degree = m\widehat {FGH} [/tex]
[tex] \huge \purple {\boxed {m\widehat {FGH} =238\degree}} [/tex]
