Question

problem 14: (first taught in lesson 49) in △ghj, m∠h = 110, m∠j = 40, and i is the point on (overline{gj}) that makes (overline{hi}) bisect ∠ghj. what is the longest side of △hij? after you enter your answer press go.

Explanation:

Step1: Find \(m\angle G\) in \(\triangle GHJ\)

By the triangle - angle sum theorem (\(m\angle G+m\angle H + m\angle J=180^{\circ}\)), substitute \(m\angle H = 110^{\circ}\) and \(m\angle J = 40^{\circ}\).
\(m\angle G=180-(110 + 40)=30^{\circ}\)

Step2: Use the angle - bisector property

Since \(HI\) bisects \(\angle GHJ\), \(m\angle IHJ=\frac{1}{2}m\angle GHJ\). So \(m\angle IHJ = 55^{\circ}\)

Step3: Find \(m\angle H IJ\) in \(\triangle HIJ\)

By the triangle - angle sum theorem (\(m\angle H IJ+m\angle IHJ + m\angle J=180^{\circ}\)), substitute \(m\angle IHJ = 55^{\circ}\) and \(m\angle J = 40^{\circ}\).
\(m\angle H IJ=180-(55 + 40)=85^{\circ}\)

Step4: Apply the side - angle relationship

In a triangle, the side opposite the largest angle is the longest side. In \(\triangle HIJ\), since \(m\angle H IJ = 85^{\circ}\), \(m\angle IHJ = 55^{\circ}\), \(m\angle J = 40^{\circ}\), and the side opposite \(\angle H IJ\) is \(HJ\)

Answer:

\(HJ\)