A ship is sailing due north. At a certain point the bearing of a lighthouse is N 42.8∘E and the distance is 10.5. After a while, the captain notices that the bearing of the lighthouse is now S 59.7∘E. How far did the ship travel between the two observations of the lighthouse?
Answers
Answer:
How far did the ship travel between the two observations of the lighthouse = 9.29
Step-by-step explanation:
the first step to answer this question is drawing the illustration as the attachment.
P is the ship, R is the light house and Q is the bearing.
PR is the distance between the ship and the light house, PR = 10.5
∠P = 42.8°, ∠Q = 59.7°
Thus, ∠R = 180° - ∠P - ∠Q
= 180° - 42.8°- 59.7°
= 77.5°
PQ is the the distance of the ship moving. We can use the sinus equation
=
=
PQ = ()(sin 59.7°)
= 9.29
Final answer:
Using trigonometric principles, the ship is estimated to have traveled approximately 19.8 miles between the two observations.
Explanation:
Your question involves the application of trigonometry in real life, in this case, calculating the distance traveled by a ship. The first sighting puts the lighthouse at N 42.8 degrees E, and the second sighting puts it at S 59.7 degrees E. So, the angle turned by the ship, relative to the lighthouse is 42.8 degrees + 59.7 degrees = 102.5 degrees.
We know the distance to the lighthouse from the first sighting is 10.5 units (let's say miles), and we need to find the distance traveled by the ship in the meantime. So, if we draw this situation it will resemble a triangle with the lighthouse as one point, and the initial and final positions of the ship as other points. The triangle will have one angle (between the initial position of the ship, the lighthouse, and the final position of the ship) of 102.5 degrees and one side (distance from the lighthouse to the initial position of the ship) of 10.5 miles. Now, the side of a triangle opposite an angle in a triangle is given by the side adjacent to the angle times the tangent of the angle.
So, the distance traveled by the ship = 10.5 * tan(102.5) = 19.8 miles approximately.
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Answers
Answers
The value of the variable 'x' using the cosine formula will be 9.4 units.
What is a right-angle triangle?
It's a form of a triangle with one 90-degree angle that follows Pythagoras' theorem and can be solved using the trigonometry function.
Trigonometric functions examine the interaction between the dimensions and angles of a triangular form.
The value of 'x' is given by the cosine of the angle ∠VWU. And the cosine of an angle is the ratio of the base and hypotenuse of the right-angle triangle. Then we have
cos 32° = 8 / x
x = 9.4
The value of the variable 'x' using the cosine formula will be 9.4 units.
More about the right-angle triangle link is given below.
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Answer:
9.4
Step-by-step explanation:
cos 32° = 8/x
x = 8/ cos 32°
= 8/ 0.8480
= 9.4
Answers
1x3x4=12 cubic meters
you then times each by four
1x4=4
3x4=12
4x4=16
you then multiply them together
4x12x16
and the answer would be what it equals
4x12x16=768 cubic meters
so the new prism's volume would be 768 cubic meters
Answers
Answer: B) 6 minutes
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Explanation:
Let's find the slope of the table. This will tell us the rate of change.
m = slope
m = (y2-y1)/(x2-x1)
m = (25250 - 26375)/(2 - 1) .... I used the first two rows
m = -1125/1
m = -1125
The slope of -1125 means the plane's altitude is decreasing at a rate of 1125 feet per minute. This is the descent speed.
Since we're told the initial altitude was 27500 ft, this means b = 27500
This means y = mx+b turns into y = -1125x+27500
x is the number of minutes and y is the altitude at time x
As a way to check this, plug in various x values of the table. You should get corresponding y values that match the table. I'll let you do this part.
------------------
Now plug in y = 16250 and solve for x
y = -1125x+27500
16250 = -1125x+27500
-1125x+27500 = 16250
-1125x = 16250-27500
-1125x = -11250
x = -11250/(-1125)
x = 10
It takes 10 minutes after the descent period has started, for the pilot to get to 16,250 ft.
Since the pilot was contacted at the x = 4 minute mark, this means 10-4 = 6 minutes is the duration needed for the pilot to go from 23,000 ft to 16,250 ft.