Question

the measures of two angles of a triangle is in the ratio 4:5 .If the sum of these two measures is equal to the measure of the third angle ,Find the measures of all the angles

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{ABC}\mathrm{be}\mathrm{the}∆. \\ \mathrm{Let}\angle \mathrm{A}=4\mathrm{x}°;\angle \mathrm{B}=5\mathrm{x}° \\ \mathrm{Now},\angle \mathrm{C}=\angle \mathrm{A}+\angle \mathrm{B}=4\mathrm{x}°+5\mathrm{x}°=9\mathrm{x}° \\ \mathrm{Now},\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°\mathbf{}\left[\mathbf{Angle}\mathbf{}\mathbf{sum}\mathbf{}\mathbf{property}\right] \\ \Rightarrow 4\mathrm{x}°+5\mathrm{x}°+9\mathrm{x}°=180° \\ \Rightarrow 18\mathrm{x}°=180° \\ \Rightarrow \mathrm{x}=10 \\ \mathrm{So},\angle \mathrm{A}=4\mathrm{x}=4\times 10=40° \\ \angle \mathrm{B}=5\mathrm{x}=5\times 10=50° \\ \angle \mathrm{C}=9\mathrm{x}=9\times 10=90° \end{gathered}$$