Question

The angles of a $△$ are $(x-40{)}^o,(x-20{)}^o$ & $(\frac{x}{2}-10)$. Find the value of $x$ also all the angles of the $△$.

Answer 1

Given angles of a triangle are $(x-40{)}^{\circ },(x-20{)}^{\circ },(\frac{x}{2}-10{)}^{\circ }$

As we know that sum of angles in a triangle is ${180}^{\circ }$

$⟹$$(x-40{)}^{\circ }+(x-20{)}^{\circ }+(\frac{x}{2}-10{)}^{\circ }={180}^{\circ }$

$⟹\frac{5x^{\circ }}{2}={250}^{\circ }$

$⟹x=100$

All angles are $(x-40{)}^{\circ }=(100-40{)}^{\circ }={60}^{\circ },(x-20{)}^{\circ }=(100-20{)}^{\circ }={80}^{\circ },(\frac{x}{2}-10{)}^{\circ }={50}^{\circ }-{10}^{\circ }={40}^{\circ }$