Question

1. if m∠abc is one degree less than three times m∠abd and m∠dbc = 47, find each measure. m∠abd = __ m∠abc = __

Explanation:

Step1: Set up the angle - relationship equation

Let \(m\angle ABD=x\). Then \(m\angle ABC = 3x - 1\). Since \(\angle ABC=\angle ABD+\angle DBC\), we have the equation \(3x - 1=x + 47\).

Step2: Solve the equation for \(x\)

Subtract \(x\) from both sides: \(3x-x-1=x - x+47\), which simplifies to \(2x-1 = 47\). Then add 1 to both sides: \(2x-1 + 1=47 + 1\), so \(2x=48\). Divide both sides by 2: \(x=\frac{48}{2}=24\).

Step3: Find \(m\angle ABD\)

Since \(x = m\angle ABD\), then \(m\angle ABD = 24^{\circ}\).

Step4: Find \(m\angle ABC\)

Substitute \(x = 24\) into the expression for \(m\angle ABC\): \(m\angle ABC=3x - 1=3\times24-1=72 - 1=71^{\circ}\).

Answer:

\(m\angle ABD = 24^{\circ}\), \(m\angle ABC = 71^{\circ}\)