Question

Find the measure of each angle in Triangle ABC. m∠a=(50−x)°m∠b=(9x−40)°m∠c=90°A. m∠a=60°m∠b=130°m∠c=90°B. m∠a=130°m∠b=60°m∠c=90°C. m∠a=40°m∠b=50°m∠c=90°D. m∠a=50°m∠b=40°m∠c=90°

Answer 1

The following sides of the triangle ABC is given below,

`\begin{gathered} m\angle a=(50-x)^0 \\ m\angle b=(9x-40)^0 \\ m\angle c=90^0 \end{gathered}`

The sum of angles in a triangle is 180°.

Therefore,

`\begin{gathered} m\angle a+m\angle b+m\angle c=180^0 \\ (50-x)^0+(9x-40)^0+90^0=180^0 \end{gathered}`
`\begin{gathered} 50^0-x^0+9x-40^0+90^0=180^0 \\ \text{collect like terms,} \\ -x^0+9x^0+50^0-40^0+90^0=180^0 \\ 8x+100^0=180^0 \\ 8x=180^0-100^0 \end{gathered}`
`\begin{gathered} 8x=80^0 \\ (8x)/(8)=(80^0)/(8) \\ x=10^0 \end{gathered}`

Let us now get the various angles,

`\begin{gathered} m\angle a=(50-x)^0=(50-10)^0=40^0 \\ m\angle b=(9x-40)^0=(9(10)-40)^0=(90-40)^0=50^0 \\ m\angle c=90^0 \end{gathered}`

Hence, the correct option is C.