Help! 三角學

1. Let h be the height of the tree.
Let the distance of H from the tree be x, the distance of L from the tree be y.
At position H,
tan H=h/x
tan 50°=h/x
x =h/ tan 50°
Similarly at position L,
tan L=h/y
tan 20°=h/y
y =h/ tan 20°
L is 40 feet farther apart from H, so
y-x=40
h/ tan 20°- h/ tan 50°=40
h(1/ tan 20°- 1/ tan 50°)=40
h=40*(1/ tan 20°- 1/ tan 50°)=20.96 feet (to 2 decimal places)

2. The given information is "SSA" (means "Side, Side, Angle"), it results in one triangle.
To solve an SSA triangle
•use The Law of Sines first to calculate one of the other two angles;
•then use the three angles add to 180° to find the other angle;
•finally use The Law of Sines again to find the unknown side.

a=6, b=2, A=20°
By The Law of Sines,
sin(A)/a = sin(B)/b
sin(20°)/6 = sin(B)/2
sin(B) = [2×sin(20°)]/6
sin(B) = 0.1140...
B = sin^−1(0.1140...)
B = 6.5463...°
B = 6.5° (to one decimal place)
C = 180°-A-B =180°-20°-6.5°=153.5°

By The Law of Sines,
sin(A)/a = sin(C)/c
sin(20°)/6 = sin(153.5°)/c
c= 6*sin(153.5°)/sin(20°)=7.84 (to 2 decimal places)