AileanaH347382 AileanaH347382
  • 30-10-2022
  • Mathematics
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given :AD is an altitude of ABC BD =3x DC =5x-14 AC =8x-10 ADC=5x+20 DAC=2x+2equation set up:__________

given AD is an altitude of ABC BD 3x DC 5x14 AC 8x10 ADC5x20 DAC2x2equation set up class=

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BrytenN614395 BrytenN614395
  • 30-10-2022

Line AD is an altitude of triangle ABC.

Angle ADC = 5x + 20

Angle ADC = 90 degrees.

Therefore;

[tex]\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}[/tex]

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