Question

1. f is in the interior of ∠apb and m∠apb=(5x + 11)°, m∠apf=(6x - 6)° and m∠fpb=(9x - 23)°. a. draw a diagram and label the angles. b. solve for x. c. find: m∠apf = m∠apb =

Explanation:

Step1: Use angle - addition postulate

Since $\angle APB=\angle APF + \angle FPB$, we can set up the equation $(5x + 11)=(6x - 6)+(9x - 23)$.

Step2: Simplify the right - hand side of the equation

Combine like terms on the right - hand side: $(6x - 6)+(9x - 23)=6x+9x-6 - 23=15x-29$. So the equation becomes $5x + 11=15x-29$.

Step3: Solve for x

Subtract $5x$ from both sides: $11 = 15x-5x-29$, which simplifies to $11 = 10x-29$. Then add 29 to both sides: $11 + 29=10x$, so $40 = 10x$. Divide both sides by 10 to get $x = 4$.

Step4: Find $m\angle APB$

Substitute $x = 4$ into the expression for $m\angle APB$: $m\angle APB=5x + 11=5\times4+11=20 + 11=31^{\circ}$.

Step5: Find $m\angle APF$

Substitute $x = 4$ into the expression for $m\angle APF$: $m\angle APF=6x - 6=6\times4-6=24 - 6=18^{\circ}$.

Answer:

a. (Diagram description): Draw a point $P$ with three rays $PA$, $PF$, and $PB$ such that $\angle APB=(5x + 11)^{\circ}$, $\angle APF=(6x - 6)^{\circ}$, and $\angle FPB=(9x - 23)^{\circ}$. Label the angles accordingly.
b. $x = 4$
c. $m\angle APB=31^{\circ}$, $m\angle APF=18^{\circ}$