Question

bd bisects \\( \angle abc \\). find the indicated measure/value.

1. find x.
2. find \\( \angle abd \\).
3. find \\( \angle bdc \\).

Explanation:

44.

Step1: Set up the equation

Since \(\overrightarrow{BD}\) bisects \(\angle ABC\), then \(\angle ABD=\angle DBC\). So \(x + 15=4x-45\).

Step2: Solve the equation

Subtract \(x\) from both sides: \(15 = 3x-45\).
Add \(45\) to both sides: \(60=3x\).
Divide both sides by \(3\): \(x = 20\).

Step1: Set up the equation

Since \(\overrightarrow{BD}\) bisects \(\angle ABC\), then \(\angle ABD=\angle DBC\). So \(2x + 35=5x-22\).

Step2: Solve the equation

Subtract \(2x\) from both sides: \(35 = 3x-22\).
Add \(22\) to both sides: \(57=3x\).
Divide both sides by \(3\): \(x = 19\).

Step3: Find \(\angle ABD\)

Substitute \(x = 19\) into \(2x+35\). \(\angle ABD=2\times19 + 35=38 + 35=73^{\circ}\).

Step1: Set up the equation

Since \(\overrightarrow{BD}\) bisects \(\angle ABC\), then \(\angle ABD=\angle DBC\). So \(10x-51=6x - 11\).

Step2: Solve the equation

Subtract \(6x\) from both sides: \(4x-51=-11\).
Add \(51\) to both sides: \(4x=40\).
Divide both sides by \(4\): \(x = 10\).

Step3: Find \(\angle BDC\)

\(\angle BDC = 180-(6x - 11)\) (linear - pair with \(\angle DBC\)). Substitute \(x = 10\), \(\angle BDC=180-(6\times10 - 11)=180-(60 - 11)=180 - 49=131^{\circ}\).

Answer:

\(x = 20\)

45.