Question

I need help on this problem

Answer 1

I've redrawn and labeled the given image. Angles
`x'` are supplementary to angles
`x`; that is, for any angle
`x`,

`x+x'=180^\circ`

It's also useful to know that for any convex
`n`-gon, the sum of its interior angles is

`(n-2)*180^\circ`

Notice that the angle supplementary to the one labeled
`b` is congruent to the angle supplementary to the one labeled
`c`; this means that
`b=c`.

So we have

`\begin{cases}b=c\\a'+e+40^\circ+90^\circ=360^\circ\\a+b'+e'=180^\circ\\a'+b+\angle(1)+d+88^\circ=540^\circ\\b'+f'+\angle(1)'=180^\circ\\e+b+f+c+g=540^\circ\\c'+f'+h'=180^\circ\\\angle(1)+f+h+76^\circ+90^\circ=540^\circ\\g'+c'+\angle(2)'=180^\circ\\a+\angle(2)+c+h+58^\circ=540^\circ\end{cases}`

Solving the system will give you

`a=61^\circ,b=130^\circ,c=130^\circ,d=59^\circ,e=111^\circ,f=86^\circ,g=83^\circ,h=144^\circ`