Math trigonometry world problems (sine law, cosine law)

Question

1. Rona is sitting in the library pondering about a mathematical concept. She states that “the cosine law can be used anytime you are given two sides and an angle” Explain the validity the statement. Please use diagrams to support your argument.

1. A boat leaves Kingston and heads in the direction of N 50o E at a speed of 50 km/h for 2.5 hours. A second boat travels in a direction of S 15° W at a speed of 70 km/h for 2.5 hours. How far are they apart from each other after 2.5 hours?  

Provide a detailed diagram to support your answer.                     

1. Johnny is standing on his apartment balcony looking across the street at a large tree. The angle of elevation from Johnny to the top of the tree is 16°and the angle of depression to the base of the tree is 41°. Johnny’s balcony is 15 m above the ground. How tall is the tree?

1. Two angles of a triangle are 115° and 45°. The longest side is 8m longer than the shortest side. Determine the length of the longest side.

Answer 1

First question

Law of Cosines, a² + b² = c² - 2(a)(b)cosθ. When you have two sides, you can figure out the remaining side from different formulas. i.e. tanθ = opp/adj, sinθ=opp/hyp, cosθ=adj/hyp.

2nd question

Let Kingston harbor be (0,0). The first boat travels 50º east of the y-axis to point b₁ for 2.5 hrs at 50km/h = 125km.

So let r₁ = 125. Let the vertical, or side adjacent to the angle θ, be y₁ and let the horizontal opposite of θ be x₁. We can determine the coordinates of point b₁ located in quadrant I

y₁ = (r₁)cos(50) = (125)cos(50) = 80.35

x₁ = (r₁)sin(50) = (125)sin(50) = 95.76

The second boat travels 15º west of the y-axis to the point b₂ for 2.5 hrs at 70km/h = 175km.

Let r₂ = 175. Let the vertical side (adjacent to the y-axis) be y₂ and the horizontal opposite of angle θ be x₂.

We can now determine the coordinates of point b₂ located in quadrant 3

y₂ = (r2)cos(15) = (175)cos(15) = 169.04

x₂ = (r2)sin(15) = (175)sin(15) = 45.3

We need to adjust our x₂ and y₂ to reflect the quadrant they're in. Both are negative so

y₂ = -169.04

x₂ = -45.3

b1(95.76, 80.35)

b2(-45.3, -169.04)

Now that we have the two coordinates, we can use the distance formula to determine how far apart they are:

d(b1,b2) = √[(x₂-x₁)²+(y₂-y₁)²] = √[(-45.3-95.76)²+(-169.04-(80.35))²] = √(19897.9+62195.4) = 286.5

The two boats are 286.5 km apart.

Third question

We have two angles and one side. First we calculate the distance from the building to the tree.

Let r = distance from Johnny to the base of the tree, or the hypotenuse of a right triangle. The angle is 41° so we can use the fact that (r)cos(41) = 15, so r = 15/cos(41)

Knowing r we can calculate the distance from the building to the tree, (r)sin(41) = x. Substituting in for r we get

(15)sin(41)/cos(41) = x = (15)tan(41) = 13.

The distance from the balcony to the tree is the same so, using the identity tanθ = opp/adj, from the second measurement we get tan(16) = y/15 or y = (15)tan(16) = 3.74.

The tree height is distance from the ground to the balcony then from the balcony to the top of the tree, or 15 + 3.74 = 18.74 meters.

Fourth question

From the given two angles we can determine the other angle to be 180-(115+45) = 20.

Using the Law of Sines we can determine the length of the longest side from the given information.

Let the shortest side be length x, so the longest side is 8+x. If we sketch the given angles we can easily put together the identities for the Law of Sines.

(x + 8)/sin(115) = x/sin(20)

(x+8)/x = sin(115)/sin(20) = 2.65

x/x + 8/x = 2.65

8/x = 2.65 - 1 = 1.65

8/1.65 = x

x = 4.85

So length of longest side is 8 + 4.85 = 12.85

Answer 2

Oblique triangle for side "c"; a^2+b^2-2ab(cosγ)=c^2.

Different formulas for diffedent sides.

n50°e; 50km/hr; 2.5hr

S 15° W at a speed of 70 km/h for 2.5

Using the same formula as above:

side "a"=125

b 185

c ?

angles "α"=?

β ?

γ 125°

You want side "c". 125^2+185^2-(2*125*185)*cos125=c^2

15625+34225-(46250)(-.5735)=c^2

49850 +26.5279 =c^2

76377.91^.5=276.365km