Question

In $△ABC,\angle A=x^o,\angle B={(2x-15)}^o$ and $\angle C={(3x+21)}^o$. Find the value of $x$ and the measure of each angle of the triangle.

• A. $x={27}^{\circ },\angle A={27}^{\circ },\angle B={41}^{\circ },\angle C={101}^{\circ }$
• B. $x={29}^{\circ },\angle A={29}^{\circ },\angle B={43}^{\circ },\angle C={108}^{\circ }$
• C. $x={27}^{\circ },\angle A={27}^{\circ },\angle B={41}^{\circ },\angle C={108}^{\circ }$
• D. $x={30}^{\circ },\angle A={30}^{\circ },\angle B={41}^{\circ },\angle C={109}^{\circ }$

Answer 1

Answer: (B)

$x={29}^{\circ },\angle A={29}^{\circ },\angle B={43}^{\circ },\angle C={108}^{\circ }$

Given, $\angle A=x^o$, $\angle B=(2x-15{)}^o$ and $\angle C=(3x+21{)}^o$.

We know, by angle sum property, the sum of angles of a triangle is ${180}^o$.

Then, $\angle A+\angle B+\angle C={180}^o$

$⟹$ $x^o+(2x-15{)}^o+(3x+21{)}^o={180}^o$

$⟹$ $6x^o+6^o={180}^o$

$⟹$ $6x^o={180}^o-6^o$

$⟹$ $6x^o={174}^o$

$⟹$ $x^o={29}^o$.

Therefore,

$\angle A=x^o={29}^o$,

$\angle B=(2x-15{)}^o=2\times {29}^o-{15}^o={58}^o-{15}^o={43}^o$

and $\angle C=(3x+21{)}^o=3\times {29}^o+{21}^o={87}^o+{21}^o={108}^o$.

Hence, option $B$ is correct.