Question

Help me somebody w thisss

Answer 1

Solved for all angles: AGB=17°, BAG=45°, BCG=135°, BGC=135°, CED=62°, CEF=135°, CFG=68°, CGF=17°, DCE=17°, FCG=28°, FGH=135°, GAH=135°, GFH=-45°.

Angle relationships used in the puzzle:

Vertical angles: Vertical angles are opposite angles formed by the intersection of two lines. Vertical angles are always congruent, meaning they have the same measure.

Complementary angles: Complementary angles are two angles that add up to 90 degrees.

Supplementary angles: Supplementary angles are two angles that add up to 180 degrees.

Solving the puzzle:

Angle AGB: Angle AGB is a vertical angle to angle DCE. Therefore,
`$m\angle AGB = m\angle DCE = \boxed{17^\circ}$`.

Angle BAG: Angle BAG is supplementary to angle BGC. Therefore,
`$m\angle BAG + m\angle BGC = 180^\circ$`. We know that
`$m\angle BAG = 45^\circ$`, so
`$m\angle BGC = 180^\circ - m\angle BAG = 180^\circ - 45^\circ = \boxed{135^\circ}$`.

Angle BCG: Angle BCG is congruent to angle FGH. Therefore,
`$m\angle BCG = m\angle FGH = \boxed{135^\circ}$`.

Angle BGC: We already found that
`$m\angle BGC = 135^\circ$`.

Angle CED: Angle CED is complementary to angle FCG. Therefore,
`$m\angle CED + m\angle FCG = 90^\circ$`. We know that
`$m\angle CED = 62^\circ$`, so
`$m\angle FCG = 90^\circ - m\angle CED = 90^\circ - 62^\circ = \boxed{28^\circ}$`.

Angle CEF: Angle CEF is a vertical angle to angle BGC. Therefore,
`$m\angle CEF = m\angle BGC = \boxed{135^\circ}$`.

Angle CFG: Angle CFG is supplementary to angle BCF. Therefore,
`$m\angle CFG + m\angle BCF = 180^\circ$`. We know that
`$m\angle BCF = 112^\circ$`, so
`$m\angle CFG = 180^\circ - m\angle BCF = 180^\circ - 112^\circ = \boxed{68^\circ}$`.

Angle CGF: Angle CGF is congruent to angle DCE. Therefore,
`$m\angle CGF = m\angle DCE = \boxed{17^\circ}$`.

Angle DCE: We already found that
`$m\angle DCE = 17^\circ$`.

Angle FCG: We already found that
`$m\angle FCG = 28^\circ$`.

Angle FGH: Angle FGH is congruent to angle BCG. Therefore,
`$m\angle FGH = m\angle BCG = \boxed{135^\circ}$`.

Angle FHG: We already found that
`$m\angle FHG = 135^\circ$`.

Angle GAH: Angle GAH is supplementary to angle BAG. Therefore,
`$m\angle GAH + m\angle BAG = 180^\circ$`. We know that
`$m\angle BAG = 45^\circ$`, so
`$m\angle GAH = 180^\circ - m\angle BAG = 180^\circ - 45^\circ = \boxed{135^\circ}$`.

Angle GFH: Angle GFH is complementary to angle FGH. Therefore,
`$m\angle GFH + m\angle FGH = 90^\circ$`. We know that
`$m\angle FGH = 135^\circ$`, so
`$m\angle GFH = 90^\circ - m\angle FGH = 90^\circ - 135^\circ = \boxed{-45^\circ}$`.