Question
$PQRSTU$ is a regular hexagon. Determine each angle of $\triangle\text{PQT}$.
✓
Answer
$\therefore$ If each interior angle $=\frac{2\text{n}-4}{\text{n}}\times90^\circ$
$=\frac{2\times6-4}{\text{6}}\times90^\circ$
$=\frac{8}{\text{6}}\times90^\circ=120^\circ$
In $\triangle\text{PUT},\text{PU}=\text{UT}$
$\angle\text{UPT}=\angle\text{UTP}$
But $\angle\text{UPT}+\angle\text{UTP}=180^\circ-\angle\text{U}$
$=180^\circ-120^\circ=60^\circ$
$\angle\text{UPT}=\angle\text{UTP}=30^\circ$
$\angle\text{TPQ}=120^\circ-30^\circ=90^\circ$ ($QT$ is diagonal which bisect $∠Q$ and $∠T$)
$\angle\text{PQT}=\frac{120}{2}=60^\circ$
Now in $\triangle\text{PQT}$,
$\angle\text{TPQ}+\angle\text{PQT}+\angle\text{PTQ}=180^\circ$ (Sum of angles of a triangle)
$\Rightarrow90^\circ+60^\circ+\angle\text{PTQ}=180^\circ$
$\Rightarrow150^\circ+\angle\text{PTQ}=180^\circ$
$\Rightarrow\angle\text{PTQ}=180^\circ-150^\circ=30^\circ$
Hence in $\triangle\text{PQT}$,
$\angle\text{P}=90^\circ$
$\angle\text{Q}=60^\circ$
and
$\angle\text{T}=30^\circ$
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