Question

in △acd, if ac = ad, a=(3x - 4)°, c=(5x + 1)°, and d=(7x - 27)°, find x and the measure of each angle

Explanation:

Step1: Recall isosceles - triangle property

Since \(AC = AD\) in \(\triangle ACD\), then \(\angle C=\angle D\).
So, \(5x + 1=7x - 27\).

Step2: Solve the equation for \(x\)

Subtract \(5x\) from both sides: \(1 = 7x-5x - 27\), which simplifies to \(1 = 2x-27\).
Add 27 to both sides: \(2x=1 + 27\), so \(2x = 28\).
Divide both sides by 2: \(x = 14\).

Step3: Find the measure of each angle

For \(\angle A\): Substitute \(x = 14\) into the expression for \(\angle A\), \(\angle A=(3x - 4)^{\circ}=(3\times14 - 4)^{\circ}=(42-4)^{\circ}=38^{\circ}\).
For \(\angle C\): Substitute \(x = 14\) into the expression for \(\angle C\), \(\angle C=(5x + 1)^{\circ}=(5\times14 + 1)^{\circ}=(70 + 1)^{\circ}=71^{\circ}\).
For \(\angle D\): Since \(\angle C=\angle D\), \(\angle D = 71^{\circ}\).

Answer:

\(x = 14\), \(\angle A=38^{\circ}\), \(\angle C = 71^{\circ}\), \(\angle D = 71^{\circ}\)