Question

In a ∆ABC, ∠A = x°, ∠B = (3x − 2)°, ∠C = y°. Also, ∠C − ∠B = 9°. Find the three angles.

Answer 1

Let , and

$$\begin{gathered} \angle C-\angle B=9^{°} \\ \Rightarrow \angle C=9^{°}+\angle B \\ \Rightarrow \angle C=9^{°}+3x^{°}-2^{°} \\ \Rightarrow \angle C=7^{°}+3x^{°} \end{gathered}$$

Substitute in above equation we get ,

$y^{°}=7^{°}+3x^{°}$

$$\begin{gathered} \angle A+\angle B+\angle C=180° \\ \Rightarrow x^{°}+(3x^{°}-2^{°})+(7^{°}+3x^{°})={180}^{°} \\ \Rightarrow 7x^{°}+5^{°}={180}^{°} \\ \Rightarrow 7x^{°}={180}^{°}-5^{°}={175}^{°} \\ \Rightarrow x^{\circ }=\frac{{175}^{\circ }}{7^{\circ }}={25}^{\circ } \\ \angle A=x^{\circ }={25}^{\circ } \\ \angle B=(3x-2{)}^{\circ }=3({25}^{\circ })-2^{\circ }={73}^{\circ } \\ \angle C=(7^{°}+3x^{°})=7^{\circ }+3(25{)}^{\circ }={82}^{\circ } \\ \angle A={25}^{\circ },\angle B={73}^{\circ },\angle C={82}^{\circ } \\ \mathrm{Hence},\mathrm{the}\mathrm{answer}. \end{gathered}$$