In ∆FGH, FH=6 ft, FG=11 ft, and m∠F=73°. Find m∠G to the nearest degree. *Hint: Law of Cosines
To find the measure of angle G, we can use the Law of Cosines. The Law of Cosines relates the lengths of the sides of a triangle to the cosine of one of its angles.
According to the Law of Cosines, we have the following formula:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case, we are given FH = 6 ft (which corresponds to side a), FG = 11 ft (which corresponds to side b), and angle F = 73° (which corresponds to angle C). We want to find the measure of angle G, which corresponds to angle G.
Let's substitute the given values into the formula:
FH^2 = FG^2 + GH^2 - 2 * FG * GH * cos(F)
6^2 = 11^2 + GH^2 - 2 * 11 * GH * cos(73°)
36 = 121 + GH^2 - 22GH * cos(73°)
Rearranging the equation:
GH^2 - 22GH * cos(73°) + 121 = 36
GH^2 - 22GH * cos(73°) + 85 = 0
Now, we can solve this quadratic equation to find the value of GH. By solving the quadratic equation, we find that GH ≈ 6.067 ft.
Now, to find the measure of angle G, we can use the Law of Sines. The Law of Sines relates the lengths of the sides of a triangle to the sines of its angles.
According to the Law of Sines, we have the following formula:
sin(A)/a = sin(B)/b = sin(C)/c
In this case, we know angle F = 73°, side FG = 11 ft (which corresponds to side b), and side GH ≈ 6.067 ft (which corresponds to side c).
Let's substitute the values into the formula:
sin(73°)/11 = sin(G)/6.067
Now, we can solve this equation for sin(G) by multiplying both sides by 6.067:
sin(G) = (sin(73°)/11) * 6.067
Using a calculator, we find that sin(G) ≈ 0.4001.
Finally, to find the measure of angle G, we can take the inverse sine (or arcsine) of sin(G):
G ≈ sin^(-1)(0.4001)
Using a calculator, we find that G ≈ 23.05°.
Therefore, the measure of angle G to the nearest degree is 23°.
To find the measure of angle G in triangle FGH, we can use the Law of Cosines. The Law of Cosines is given by the equation:
c^2 = a^2 + b^2 - 2ab * cos(C)
where c represents the side opposite angle C.
In our problem, FH is side c, FG is side b, and m∠F is angle C. Let's plug in the values we have:
FH = 6 ft
FG = 11 ft
m∠F = 73°
Using the Law of Cosines, we have:
FH^2 = FG^2 + GH^2 - 2 * FG * GH * cos(m∠G)
Substituting the known values:
6^2 = 11^2 + GH^2 - 2 * 11 * GH * cos(m∠G)
36 = 121 + GH^2 - 22GH * cos(m∠G)
Rearranging the equation:
GH^2 - 22GH * cos(m∠G) + 85 = 0
Now we have a quadratic equation. To solve it, we need to use the quadratic formula:
GH = (-b ± √(b^2 - 4ac)) / 2a
In our case:
a = 1
b = -22 * cos(m∠G)
c = 85
Let's solve for GH:
GH = (-(-22 * cos(m∠G)) ± √((-22 * cos(m∠G))^2 - 4 * 1 * 85)) / (2 * 1)
Simplifying:
GH = (22 * cos(m∠G) ± √(484 * cos^2(m∠G) - 340)) / 2
To find GH, the length of side GH, we need to choose the positive root.
Now, using the information given, we can find GH.
well, using the law of cosines,
f^2 = 6^2 + 11^2 - 2(6)(11) cos73°
f = 10.88
now, using the law of sines,
sinG/6 = sin73°/10.88