Question

refer to the figure. if $mangle2=(a + 15)^{circ}$ and $mangle3=(a + 35)^{circ}$, find the value of a such that $overrightarrow{hl}perpoverrightarrow{hj}$. a =

Explanation:

Step1: Recall perpendicular - angle property

If $\overrightarrow{HL}\perp\overrightarrow{HJ}$, then $\angle JHL = 90^{\circ}$, so $m\angle2 + m\angle3=90^{\circ}$.

Step2: Substitute angle measures

Substitute $m\angle2=(a + 15)^{\circ}$ and $m\angle3=(a + 35)^{\circ}$ into the equation $m\angle2 + m\angle3 = 90^{\circ}$. We get $(a + 15)+(a + 35)=90$.

Step3: Simplify the left - hand side

Combine like terms: $a+a+15 + 35=90$, which simplifies to $2a+50 = 90$.

Step4: Solve for a

Subtract 50 from both sides: $2a=90 - 50$, so $2a = 40$. Then divide both sides by 2: $a=\frac{40}{2}=20$.

Answer:

$20$