Question

In $∆ABC,\angle A=x^{\circ },\angle B={\left(3x-2\right)}^{\circ },\angle C=y^{\circ }\mathrm{and}\angle C-\angle B=9^{\circ }.$ Find the three angles.

Answer 1

$$\begin{gathered} \because \angle C-\angle B=9^{\circ } \\ \therefore y^{\circ }-{\left(3x-2\right)}^{\circ }=9^{\circ } \\ \Rightarrow y^{\circ }-3x^{\circ }+2^{\circ }=9^{\circ } \\ \Rightarrow y^{\circ }-3x^{\circ }=7^{\circ }.....\left(i\right) \end{gathered}$$
The sum of all the angles of a triangle is ${180}^{\circ }$, therefore
$$\begin{gathered} \angle A+\angle B+\angle C={180}^{\circ } \\ \Rightarrow x^{\circ }+{\left(3x-2\right)}^{\circ }+y^{\circ }={180}^{\circ } \\ \Rightarrow 4x^{\circ }+y^{\circ }={182}^{\circ }.....\left(ii\right) \end{gathered}$$
Subtracting (i) from (ii), we have
$$\begin{gathered} 7x^{\circ }={182}^{\circ }-7^{\circ }={175}^{\circ } \\ \Rightarrow x^{\circ }={25}^{\circ } \end{gathered}$$
Now, substituting $x^{\circ }={25}^{\circ }$ in (i), we have
$y^{\circ }=3x^{\circ }+7^{\circ }=3\times {25}^{\circ }+7^{\circ }={82}^{\circ }$
Thus
$$\begin{gathered} \angle A=x^{\circ }={25}^{\circ } \\ \angle B={\left(3x-2\right)}^{\circ }={75}^{\circ }-2^{\circ }={73}^{\circ } \\ \angle C=y^{\circ }={82}^{\circ } \end{gathered}$$
Hence, the angles are ${25}^{\circ },{73}^{\circ }\mathrm{and}{82}^{\circ }$.