Question

find the value of b such that hl ⊥ hj. in the figure, m∠2=(b + 19)° and m∠3=(b + 11)°. b = □

Explanation:

Step1: Use the property of perpendicular lines

Since \(HL\perp HJ\), then \(\angle 2+\angle 3 = 90^{\circ}\) (the sum of angles that form a right - angle is \(90^{\circ}\)).

Step2: Substitute the given angle expressions

Given \(m\angle 2=(b + 19)^{\circ}\) and \(m\angle 3=(b + 11)^{\circ}\), we substitute into the equation \(\angle 2+\angle 3=90^{\circ}\). So \((b + 19)+(b + 11)=90\).

Step3: Simplify the left - hand side of the equation

Combine like terms: \(b + 19+b + 11=2b+30\). So the equation becomes \(2b + 30=90\).

Step4: Solve for \(b\)

Subtract 30 from both sides: \(2b+30 - 30=90 - 30\), which gives \(2b=60\). Then divide both sides by 2: \(b=\frac{60}{2}=30\).

Answer:

\(30\)