Question

Two adjacent angles of a parallelogram are $(3x-4{)}^{\circ }and(3x+16{)}^{\circ }$. Find the value of x and hence find the measure of each of its angles.

Answer 1

In || gm ABCD, $\angle A\angle B$ are two adjacent angles

Let $\angle A=(3x-4{)}^{\circ }and\angle B=(3x+16{)}^{\circ }$

But $\angle A+\angle B={180}^{\circ }$

$\Rightarrow (3x-4{)}^{\circ }+(3x+16)={180}^{\circ }\Rightarrow 3x-4^{\circ }+3x+16={180}^{\circ }\Rightarrow 6x+{12}^{\circ }={180}^{\circ }\Rightarrow 6x={180}^{\circ }-{12}^{\circ }\Rightarrow 6x=168\Rightarrow x=\frac{168}{6}={28}^{\circ }\therefore x={28}^{\circ }Now\angle A=3x-4=3\times {28}^{\circ }-4^{\circ }={84}^{\circ }-4^{\circ }={80}^{\circ }\angle B=3x+16=3\times 28+16={84}^{\circ }+{16}^{\circ }={100}^{\circ }But\angle C=\angle A\text{opposite angles of}||gm\therefore \angle C={80}^{\circ }Similarly\angle D=\angle B={100}^{\circ }Hence\angle A={80}^{\circ },\angle B={100}^{\circ },\angle C={80}^{\circ }and\angle D={100}^{\circ }$