Respuesta :
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 307.35°
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;
[tex]\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}[/tex]
Therefore;
[tex]\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}[/tex]
Which gives;
[tex]\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}[/tex]
[tex]\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}[/tex]
Similarly, we can get;
[tex]\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}}[/tex]
The angle subtended by the distance of the buoy from the port, C 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° ≈ 307.35°
Learn more about bearing in mathematics here:
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