Question

given :AD is an altitude of ABC BD =3x DC =5x-14 AC =8x-10 ADC=5x+20 DAC=2x+2equation set up:__________

Answer 1

Line AD is an altitude of triangle ABC.

Angle ADC = 5x + 20

Angle ADC = 90 degrees.

Therefore;

`\begin{gathered} 5x+20=90 \\ 5x=90-20 \\ 5x=70 \\ \text{Divide both sides by 5} \\ x=14 \\ \angle DAC=2x+2 \\ \angle DAC=2(14)+2 \\ \angle DAC=28+2 \\ \angle DAC=30 \\ \text{Therefore, the given sides are;} \\ BD=3(14) \\ BD=42 \\ DC=5x-14 \\ DC=5(14)-14 \\ DC=70-14 \\ DC=56 \\ AC=8(14)-10 \\ AC=112-10 \\ AC=102 \end{gathered}`