Question

hw 1.3.4 angles formed by intersecting lines
m < hqp =
∠eqh ≅ ∠pqf
m∠eqh = 6x + 18, m∠eqf = x² + x, m∠hqf = 2x + 12
find m∠hqp.

Explanation:

Step1: Use angle - relationship

Since $\angle EQH\cong\angle PQF$, and $\angle EQF=\angle EQH+\angle HQF$, we have the equation $x^{2}+x=(6x + 18)+(2x + 12)$.

Step2: Simplify the equation

$x^{2}+x=6x + 18+2x + 12$, which simplifies to $x^{2}+x=8x + 30$. Rearrange it to the standard quadratic - form: $x^{2}+x-8x - 30 = 0$, so $x^{2}-7x - 30 = 0$.

Step3: Factor the quadratic equation

Factor $x^{2}-7x - 30$ as $(x - 10)(x+3)=0$. Then $x - 10 = 0$ or $x + 3 = 0$, so $x = 10$ or $x=-3$. Since angles cannot be negative in this context, we take $x = 10$.

Step4: Find the measure of $\angle EQH$

Substitute $x = 10$ into the expression for $\angle EQH$: $m\angle EQH=6x + 18=6\times10 + 18=60 + 18 = 78^{\circ}$.

Step5: Find the measure of $\angle HQF$

Substitute $x = 10$ into the expression for $\angle HQF$: $m\angle HQF=2x + 12=2\times10+12=20 + 12 = 32^{\circ}$.

Step6: Find the measure of $\angle HQP$

Since $\angle EQH\cong\angle PQF$, and $\angle HQP=\angle EQH+\angle HQF$, then $m\angle HQP=78^{\circ}+32^{\circ}=110^{\circ}$.

Answer:

$110^{\circ}$