In ΔFGH, FH = 9 ft, FG = 11 ft, and m∠F = 65°. Find m∠G. Round your answer to the nearest tenth.

Question

A. 41.4°
B. 10.9°
C. 48.6°
D. 89.7°

Answer 1

Using the Law of Cosines, we have:

$GH^2 = FG^2 + FH^2 - 2(FG)(FH)\cos(\angle F)$

Substituting the given values:

$GH^2 = 11^2 + 9^2 - 2(11)(9)\cos(65^\circ)$

Solving for $GH$:

$GH \approx 6.4$

Now, we can use the Law of Sines to find $\angle G$:

$\frac{FG}{\sin(\angle G)} = \frac{GH}{\sin(\angle F)}$

Substituting the values we know:

$\frac{11}{\sin(\angle G)} = \frac{6.4}{\sin(65^\circ)}$

Solving for $\angle G$:

$\angle G \approx 41.4^\circ$

So the answer is (A) 41.4°.

Answer 2

AAAaannndd the bot gets it wrong yet again!

11^2 + 9^2 - 2(11)(9)cos65° so GH = 10.877
sinG9 = sin65°/10.877