Question

Construct $\Delta$ABC, given $m\angle A={60}^o$, $m\angle B={30}^o$ and $AB=5.8$ cm.

Answer 1

Construct triangle $ABC$

$m\angle A=60,m\angle B={30}^o$ & $AB=5.8cm$

$\to$ for triangle $ABC$

$\therefore m\angle A+\angle B+m\angle C={180}^o$

$\therefore {60}^o+{30}^o+m\angle C={180}^o$

$\therefore m\angle C={180}^o-{60}^o-{30}^o$

$\therefore m\angle C={90}^o$

$\to$ so, the $△ABC$ is right angle triangle

$\to$ First draw a line $AB=5.8cm$

$\to$ using rounder make an angle $A={60}^o$ for constant ${60}^o$ draw a $r$ radius circle which intersect the line $AB$, the $AB$ us centres and same radius draw are on the drawing on the chord,

A point of centre draw aline from $A$

$\to$ using rounder, make are angle $B={30}^o$

for constant ${30}^o$ follow ${60}^o$ angle step after bisect the angle & draw a line from point $B$

$\to$ A point of contact named $C$ so, $AC=3cm,BC=5cm,m\angle c={90}^o$