Question

PQRSTU is a regular hexagon. Determine each angle of $△PQT$.

Answer 1

In regular hexagon, PQRSTU, diagonals PT and QT are joined.



$\therefore$ If each interior angle $=\frac{2n-4}{n}\times {90}^{\circ }$
$=\frac{2\times 6-4}{6}\times {90}^{\circ }=\frac{8}{6}\times {90}^{\circ }={120}^{\circ }$
In $△PUT,PU=UT\therefore \angle UPT=\angle UTP$
But $\angle UPT+\angle UTP={180}^{\circ }-\angle U={180}^{\circ }-{120}^{\circ }={60}^{\circ }\therefore \angle UPT=\angle UTP={30}^{\circ }\therefore \angle TPQ={120}^{\circ }-{30}^{\circ }={90}^{\circ }$
$(\because$ QT is diagonal which bisect $\angle Q$ and $\angle T$ )
$\therefore \angle PQT=\frac{{120}^{\circ }}{2}={60}^{\circ }$
Now in $△PQT,$
$\angle TPQ+\angle PQT+\angle PTQ={180}^{\circ }$
(sum of angles of a triangle)
$\Rightarrow {90}^{\circ }+{60}^{\circ }+\angle PTQ={180}^{\circ }\Rightarrow {150}^{\circ }+\angle PTQ={180}^{\circ }\Rightarrow \angle PTQ={180}^{\circ }-{150}^{\circ }={30}^{\circ }$
Hence in $△PQT,$
$\angle ={90}^{\circ },\angle Q={60}^{\circ }$ and $\angle T={30}^{\circ }$