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In /_\BCD,m/_B=(8x+12)^(@),m/_C=(3x-17)^(@), and m/_D=(x+17)^(@). Find m/_D. Answer:

In △BCD,m∠B=(8x+12)∘,m∠C=(3x−17)∘ \triangle \mathrm{BCD}, \mathrm{m} \angle B=(8 x+12)^{\circ}, \mathrm{m} \angle C=(3 x-17)^{\circ} , and m∠D=(x+17)∘ \mathrm{m} \angle D=(x+17)^{\circ} . Find m∠D \mathrm{m} \angle D .\newlineAnswer:

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Q. In △BCD,m∠B=(8x+12)∘,m∠C=(3x−17)∘ \triangle \mathrm{BCD}, \mathrm{m} \angle B=(8 x+12)^{\circ}, \mathrm{m} \angle C=(3 x-17)^{\circ} , and m∠D=(x+17)∘ \mathrm{m} \angle D=(x+17)^{\circ} . Find m∠D \mathrm{m} \angle D .\newlineAnswer:
  1. Recognize Triangle Angle Sum: Recognize that the sum of the angles in any triangle is 180180 degrees.
  2. Write Equation for Triangle BCD: Write an equation that represents the sum of the angles in triangle BCD.\newlinem/_B+m/_C+m/_D=180m/\_B + m/\_C + m/\_D = 180 degrees\newline(8x+12)+(3x−17)+(x+17)=180(8x + 12) + (3x - 17) + (x + 17) = 180
  3. Combine Like Terms: Combine like terms in the equation.\newline8x+12+3x−17+x+17=1808x + 12 + 3x - 17 + x + 17 = 180\newline8x+3x+x+12−17+17=1808x + 3x + x + 12 - 17 + 17 = 180\newline12x+12=18012x + 12 = 180
  4. Isolate Terms with x: Subtract 1212 from both sides of the equation to isolate the terms with xx.\newline12x+12−12=180−1212x + 12 - 12 = 180 - 12\newline12x=16812x = 168
  5. Solve for x: Divide both sides of the equation by 1212 to solve for x.\newline12x12=16812\frac{12x}{12} = \frac{168}{12}\newlinex=14x = 14
  6. Find Measure of Angle D: Substitute the value of xx back into the expression for m/_Dm/\_D to find its measure.\newlinem/_D=(x+17)m/\_D = (x + 17)\newlinem/_D=(14+17)m/\_D = (14 + 17)\newlinem/_D=31m/\_D = 31 degrees

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