Question

In $\Delta PQR,\angle Q={35}^0,\angle R={61}^0$ and the bisector of $\angle QPR$ meet $QR$ at $x$. Then arrange the sides $PX,QX$ and $RX$ in descending order of their length.

Answer 1

Given $\angle Q={35}^0$ and $\angle R={61}^0$

By angle sum property of triangle,

$\angle P+\angle Q+\angle R=180$

$\therefore \angle P+35+61=180$

$\therefore \angle P+96=180$

$\therefore \angle P={84}^0$

Segment $PX$ is bisector of $\angle P$

$\therefore \angle QPX=\angle RPX=\frac{84}{2}={42}^0$

In $△QPX$,

$\angle QPX>\angle PQX$

$\therefore QX>PX$ (1)

In $△RPX$,

$\angle PRX>\angle RPX$

$\therefore PX>RX$ (2)

From equations (1) and (2), we get,

$\therefore QX>PX>RX$