Question

1. A buoy floats 19 yards from the eastern most point of a boat and 15 yards from the western most point of a second boat. The angle formed from the buoy to the two boats 108o. What is the other angle measure that is formed with the buoy? (Law of Sines or Law of Cosines Problem)
2. A helicopter flies between two islands that are 20 miles apart. The angle of elevation with one island is 15o and 35o with the second island. What is the mileage of the helicopter? Round your answer to the nearest tenth of a mile. (Law of Sines or Law of Cosines Problem)

Answer 1

The laws of cosines and law of sines can be used given that two sides

and an included angle, or two angles a side are known.

Response:

1. The other angles in the triangle formed by the buoy are approximately;

• 31.1° and 40.9°

2. Distance of the helicopter from the first island is approximately;

• 14.5 miles

How is the Law of Sines and Cosines used?

Given parameters are;

Distance of the buoy from the easternmost point of a boat = 19 yards

Distance of the buoy from the westernmost point of the other boat = 15 yards

Angle formed from the buoy to the two boats = 108°

Distance between the two boats, d, is given by the law of cosines, as follows;

d² = 19² + 15² - 2 × 19 × 15 × cos(108°) = 586 - 570·cos(108°)

d = √(586 - 570·cos(108°))

By the law of Sines, we have;

$\dfrac{d}{sin(108^{\circ})} = \mathbf{\dfrac{15}{sin(Angle \ formed \ from \ the \ boat \ on \ the \ West, \ \theta_1)}}$

Which gives;

$sin(\theta_1) = \mathbf{ \dfrac{15 \times sin(108^{\circ})}{\sqrt{586 - 570 \cdot cos(108^{\circ})} }}$

The o

$\theta_1 = arcsin \left( \dfrac{15 \times sin(108^{\circ})}{\sqrt{586 - 570 \cdot cos(108^{\circ})} } \right) \approx \mathbf{31.1^{\circ}}$

The other angles formed in the triangle containing the buoy are;

• θ₁ ≈ 31.1

• θ₂ ≈ 180° - 108° - 31.1° ≈ 40.9°

2. Distance between the two islands = 20 miles

Angle of elevation with one island = 15°

Angle of elevation with the second island = 35°

Required:

The mileage (distance travelled) of the helicopter.

Solution:

Let A represent the island that has an angle of elevation to the helicopter

of 15°, and let B represent the other island.

Angle formed by the helicopter and the two island, θ, is found as follows;

θ = 180° - (15° + 35°) = 130°

By the Law of Sines, we have;

$\dfrac{20}{sin(130^{\circ})} = \mathbf{ \dfrac{Distance \ from \ island \ A }{sin(35^{\circ})}}$

Which gives;

$Distance \ of \ helicopter \ from \ island \ A = \mathbf{ \dfrac{20}{sin(130^{\circ})} \times sin(35^{\circ})}$

$Mileage \ from \ island \ A = \dfrac{20}{sin(130^{\circ})} \times sin(35^{\circ}) \times cos(15^{\circ}) \approx 14.5$

• The mileage of the helicopter from the first island is approximately 14.5 miles

Learn more about the Law of Sines and Cosines here:

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