Question

in the figure, ( mangle 2=(b + 11)^{circ} ) and ( mangle 3=(b + 29)^{circ} ).
find the value of ( b ) such that ( overrightarrow{hl}perpoverrightarrow{hj} ).
( b=square )

Explanation:

Step1: Use the perpendicular condition

If \(\overrightarrow{HL}\perp\overrightarrow{HJ}\), then \(m\angle JHL = 90^{\circ}\), so \(m\angle2 + m\angle3=90^{\circ}\).
Given \(m\angle2=(b + 11)^{\circ}\) and \(m\angle3=(b + 29)^{\circ}\), we substitute into the equation: \((b + 11)+(b + 29)=90\).

Step2: Simplify the left - hand side of the equation

Combine like terms: \(b+11 + b+29=(b + b)+(11 + 29)=2b+40\). So the equation becomes \(2b + 40=90\).

Step3: Solve for \(b\)

Subtract 40 from both sides: \(2b+40-40=90 - 40\), which gives \(2b=50\).
Divide both sides by 2: \(b=\frac{50}{2}\).

Answer:

\(25\)