The value of the angle subtended by the distance of the buoy from the
port is given by sine and cosine rule.
- The bearing of the buoy from the is approximately <u>307.35°</u>
Reasons:
Location from which the ship sails = Port
The speed of the ship = 10 mph
Direction of the ship = N35°W
Location of the warning buoy = 5 miles north of the port
Required: The bearing of the warning buoy from the ship after 7.5 hours.
Solution:
The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles
By cosine rule, we have;
a² = b² + c² - 2·b·c·cos(A)
Where;
a = The distance between the ship and the buoy
b = The distance between the ship and the port = 75 miles
c = The distance between the buoy and the port = 5 miles
Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°
Which gives;
a = √(75² + 5² - 2 × 75 × 5 × cos(35°))
By sine rule, we have;
![\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Ba%7D%7Bsin%28A%29%7D%20%3D%20%5Cmathbf%7B%20%5Cfrac%7Bb%7D%7Bsin%28B%29%7D%7D)
Therefore;
![\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20sin%28B%29%3D%20%5Cfrac%7Bb%20%5Ccdot%20sin%28A%29%7D%7Ba%7D)
Which gives;
![\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20sin%28B%29%20%3D%20%5Cmathbf%7B%5Cfrac%7B75%20%5Ccdot%20sin%2835%5E%7B%5Ccirc%7D%29%7D%7B%5Csqrt%7B75%5E2%20%2B%205%5E2%20-%202%20%5Ctimes%2075%5Ctimes5%5Ctimes%20cos%2835%5E%7B%5Ccirc%7D%29%20%7D%20%7D%7D)
![\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20B%20%3D%20arcsin%5Cleft%28%20%5Cfrac%7B75%20%5Ccdot%20sin%2835%5E%7B%5Ccirc%7D%29%7D%7B%5Csqrt%7B75%5E2%20%2B%205%5E2%20-%202%20%5Ctimes%2075%5Ctimes5%5Ctimes%20cos%2835%5E%7B%5Ccirc%7D%29%20%7D%20%7D%5Cright%29%20%5Capprox%2037.32%5E%7B%5Ccirc%7D)
Similarly, we can get;
![\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20B%20%3D%20arcsin%5Cleft%28%20%5Cfrac%7B75%20%5Ccdot%20sin%2835%5E%7B%5Ccirc%7D%29%7D%7B%5Csqrt%7B75%5E2%20%2B%205%5E2%20-%202%20%5Ctimes%2075%5Ctimes5%5Ctimes%20cos%2835%5E%7B%5Ccirc%7D%29%20%7D%20%7D%5Cright%29%20%5Capprox%20%5Cmathbf%7B%20142.68%5E%7B%5Ccirc%7D%7D)
The angle subtended by the distance of the buoy from the port, <em>C</em> is therefore;
C ≈ 180° - 142.68° - 35° ≈ 2.32°
By alternate interior angles, we have;
The bearing of the warning buoy as seen from the ship is therefore;
Bearing of buoy ≈ 270° + 35° + 2.32° ≈ <u>307.35°</u>
Learn more about bearing in mathematics here:
brainly.com/question/23427938