Question

Using the informations given in the adjoining figure, write the sides of triangle BOC in ascending order of length.

Answer 1

Since $BAD$ is a straight line,
$\angle BAC={180}^o-{135}^o={45}^o$
From the figure,
$\angle ABO=OBC=x$ (say)
Since $BAD$ is a straight line,
$x+x+{95}^o={180}^o$
$=>x={42.5}^o$
$=>\angle C=2x={85}^o$

From the figure, $\angle AOC=OCB=y$ (say)
$=>\angle B=2y$
Since
sum of angles in a triangle ABC $={180}^o$
$\angle A+\angle B+\angle C={180}^o$
${45}^o+2y+{85}^o={180}^o$
$=>y={50}^o$
Now, since
sum of angles in a triangle OBC $={180}^o$
$\angle OBC+\angle BOC+\angle BCO={180}^o$
$=>{50}^o+\angle BOC+{42.5}^o={180}^o$
$=>\angle BOC={87.5}^o$
Now, in triangle OBC, $\angle BOC>\angle OBC>\angle BCO$
This
means, the sides opposite to the angles will also have the same relation.
$=>BC>OC>OB$