In the figure below,
ABCD and CEFG are rhombuses.
ADH and ECD are straight lines;
| BE | = CE and CD = CG. |
| $∠CGD$ | = $71°$ |
Find $∠CBE$, $∠CGF$, and $∠GDH$.
Ans: $∠CBE$ : ______________${}^{\circ}$ [1]
Ans: $∠CGF$ : ______________${}^{\circ}$ [1]
Ans: $∠GDH$ : ______________${}^{\circ}$ [1]
| Since BE | = CE and CD = CG, and ABCD and CEFG are rhombuses, |
| CE | = CG = CD = BC, |
So triangle BCE is an equilateral triangle.
| $∠CBE$ | = $\boxed{60°}$ |
| $∠CDG$ | = $∠CGD$ = $71°$ (Angles of isosceles triangle) |
| $∠DCG$ | = $180° - 71° - 71°$ (Angles of a triangle) |
| = $38°$ |
Since ECD is a straight line,
| $∠ECG$ | = $180° - 38°$ (Angles on a straight line) |
| = $142°$ | |
| $\Rightarrow$ $∠CGF$ | = $\dfrac{1}{2}$ $\times$ $(360° - 142° - 142°)$ (Angles of a rhombus) |
| = $\boxed{38°}$ | |
| $∠BCE$ | = $60°$ (Angles of equilateral triangle) |
| $∠BCD$ | = $180° - 60°$ |
| = $120°$ | |
| $∠CDA$ | = $\dfrac{1}{2}$ $\times$ $(360° - 120° - 120°)$ (Angles of a rhombus) |
| = $60°$ |
Since ADH is a straight line,
| $∠GDH$ | = $180° - 60° - 71°$ (Angles on a straight line) |
| = $\boxed{49°}$ |
