Question

Construct a $\Delta ABC$ in which $BC=3.6$ cm,$(AC-AB)=1.6$ cm and $\angle ACB={30}^{\circ }$

Answer 1

Given : $ABC$ triangle, $\angle ACB={30}^{\circ },BC=3.6$ cm, $AC-AB=1.6$ cm
To construct : $\Delta ABC$
Steps of construction.
1. Draw a straight line $BC=3.6$ cm.
2. At $C$, make $\angle BCX={30}^{\circ }$
3. At $CX$, cut off $CD=1.6$ cm.
4. Join $BD$.
5. AT $B$, make $\angle DBY=\angle BDX$.
6. Let $BY$ cuts $DX$ at $A$.
Hence, $\Delta ABC$ is the required triangle.