Question

1. in △abc, m∠a = 40°, m∠b = 75°, and m∠c = 65°. which list has the sides of △abc in order from longest to shortest? (1) ac > bc > ab (2) bc > ab > ac (3) ab > ac > bc (4) ac > ab > bc
2. in △fgh, m∠g = m∠h, fg = x + 40, gh = 3x - 20, and fh = 2x + 20. determine the length of fh. (1) 20 (2) 40 (3) 60 (4) 80

Explanation:

Step1: Recall angle - side relationship in a triangle

In a triangle, the side opposite the largest angle is the longest and the side opposite the smallest angle is the shortest.

Step2: Compare the angles in $\triangle ABC$

Given $m\angle A = 40^{\circ}$, $m\angle B=75^{\circ}$, $m\angle C = 65^{\circ}$. Since $75^{\circ}>65^{\circ}>40^{\circ}$, the side opposite $\angle B$ is the longest, the side opposite $\angle C$ is the second - longest, and the side opposite $\angle A$ is the shortest. The side opposite $\angle A$ is $BC$, the side opposite $\angle B$ is $AC$, and the side opposite $\angle C$ is $AB$. So $AC>BC>AB$.

Step3: Solve for $x$ in $\triangle FGH$

Since $m\angle G=m\angle H$, $\triangle FGH$ is isosceles and $FG = FH$. Set up the equation $x + 40=2x+20$.
Subtract $x$ from both sides: $40=x + 20$.
Subtract 20 from both sides: $x=20$.

Step4: Find the length of $FH$

Substitute $x = 20$ into the expression for $FH$. $FH=2x + 20=2\times20+20=40 + 20=60$.

Answer:

1. (1) $AC>BC>AB$
2. (3) 60