5= 5= \,\,\frac{x}{3} 3 x

A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th

Leno4ka [110]

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;

$\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}$

Therefore;

$\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}$

Which gives;

$\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}$

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}$

Similarly, we can get;

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}$

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:

brainly.com/question/23427938

5 0

2 years ago

A line has a slope of -1/2 and a y-intercept of –2. What is the x-intercept of the line? –4, –1, 1 ,4

grin007 [14]

To get the x-intercept, we simply set y = 0 and solve for "x".

now, we have the slope and the y-intercept, well, let's plug those two in the slope-intercept form, reason why is called that anyway,

$\bf y=\stackrel{slope}{-\cfrac{1}{2}}x\stackrel{y-intercept}{-2}\implies 0=-\cfrac{x}{2}-2\implies \cfrac{x}{2}=-2\implies x=-4$

5 0

3 years ago

If a rock falls from a height of 40 meters on Earth, the height H (in meters) after x seconds is approximately

Julli [10]

Once the rock hits the ground, we know that its height will be equal to 0m
because of this, we can replace H(x) with 0
$40 - 4.9 {x}^{2} = 0$
now all we have to do is get x by itself
so..
$4.9 {x}^{2} = 40$
then
${x}^{2} = 40 \div 4.9$
${x}^{2} = 8.1632...$
now take the square root of each side
$x = 2.857...$
2.857 seconds (rounded to the third decimal)

4 0

3 years ago

Please helpppppppppppppppppppppppppppppppp

Crank

Answer:

Step-by-step explanation:

6 0

3 years ago

Need help with problem urgently

antiseptic1488 [7]

Answer:

i think it c because it hits 2 and 6 and when you divide it you get 1/3

3 0

3 years ago

5= 5= \,\,\frac{x}{3} 3 x ​

5= 5= \,\,\frac{x}{3} 3 x