Long 發問於 科學及數學數學 · 6 年前

Help! 三角學

有沒有數學三角學高手?

I need help.

Here are my questions

1. Pat needs to determine the height of a tree before cutting it down to be sure that it will not fall on a nearby fence. The angle of elevation of the tree from one position on a flat path from the tree is H=50°, and from a second position L=40 feet farther along this path it is B=20°. What is the height of the tree?

2.Two sides and an angle are given below. Determine whether the given information results in one triangle, two triangles, or no triangle at all. Solve any triangle(s) that results.

a=6,

b=2,

A=20°

1 個解答

評分
  • 6 年前
    最愛解答

    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)

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