From a lighthouse window 30.0m above the sea level, the angles of deppression of two buoys are 15 and 26 degrees respectively. To the nearest tenth of a meter, the distance between the two buoys is___
5 years ago
Answered By Vinitkumar p
ans is 6.60.
5 years ago
Answered By Vinitkumar p
suppose the first buoy whose angle of depression is 15° is x meter from light house and another one is Y meter from light house.
tan15= x/30
tan26 = y/30
y-x =6.6 meter
I hope you get the drawing by yourself
5 years ago
Answered By Daunte S
To start this question you should draw a diagram to simplify it and easily determine what needs to be measured. From the diagram below, it is clear that we must find the difference in the length of the bottom side of that triangle based on the angles we are given. The angles given are termed "angles of depression," which means that it is measured starting from looking straight ahead (as if you are on a ledge), and then looking down. Ex: an angle of depression of 15 degrees means that you looked 15 degrees down from eye level.
Now we can begin calculations. The graph shows the relation between the angles to each other. Applying the "Tangent Formula" to both triangles (tanθ = opposite/adjacent), we can find the two side lengths needed. Triangle "a" uses tan(15) = 30/adjacent, rearranged this gives us adjacent = 30/tan(15), which equals 111.96m. Applying the same process to triangle "b" gives us an adjacent length of 61.51m.
So now that we have both lengths, we need to find the difference between which is just a simple subtraction of 111.96 - 61.51 = 50.45m. Since the answer needs to be to the nearest tenth of a meter, the answer is actually 50.5m.
The whiteboard playback seems to be glitched, but examples of this problem can be found online for further clarification. The trick to this question was that the angle of depression cannot be used directly into the formula from the observers point of view which was the error that the person above me made.
For any future trigonometry questions, always try to use a diagram, thinking with shapes makes it much easier.
5 years ago
Answered By Mat B
To answer this question, it helps to know what an angle of depression is. from a horizontal line, (or someone's gaze in a straight line at eye level) an angle of depression opens downward from the horizontal line. So if someone was looking out the lighthouse window straight forward at eye level, it would be a horizontal gaze 30 meters above sea level. If the person looked down towards the buoys, their switch from a horizontal gaze to a downward gaze of 15 or 26 degrees is what creates the angle of depression.
Notice how the light house wall on the left with the window at the top makes 90 degree angles with the eye level of the person in the light house window. It looks like the 30 meter tall light house also forms a right angle with the horizontal lines at sea level that reach out to the buoys. So there are a few right angle triangles to work with here.
The distance we are solving for is between the two buoys down at sea level. Well there are triangles formed by the 30 meter verticle wall of the light house, the line of sight from the light house window down to the buoys, and the horizontal line at sea level going out to the buoys.
So if we use trigonometry some of this can form a solvable equation. We need to work with angles of the triangles formed by the light house the buoys and so on, and we need to use a side length if we can find one. We've got a 3 meter tall window that's part of the triangle so let's try to use that. We don't know the angle from the buoys back to the light house, and from the buoys to the light house window, but what else is there?
How about the angles of depression? There's a right angle formed by the 30 meter vertical wall and the horizontal sight line from the window. Right angles are 90 degrees, and we can find the measure of the angle between the light house wall and the line from the window to the buoys by subtracting 90-15, and 90-26. Those angles will be 75 degrees, and 64 degrees.
Now that we have the measure of one angle and one side, we can start solving for the distances at sea level from the light house to the buoys. Our trig expressions are Soh Cah Toa. Using the angles we just calculated, the 30 meters would be adjacent and the horizontal line at sea level to the buoys would be opposite of the angles.
So Tan75=op/30
to get the "opp" part by itself in the equation, we multiply both sides by 30 and we see
30Tan(75)=opp
plugging that into your calculator gives 111.96 which is the distance to the furthest buoy. We need to subtract the distance of the light house to the furthest buoy (111.96) from that of the light house to the closer buoy to get the distance between the two buoys.
Using the same trig expression can get us closer to the solution. So using the 64 degree angle and the 30 meter measurement we can start manipulating the Tan expression
Tan64= opp/30
and multiply both sides to solve for the opp side
30Tan(64)=opp = 61.51
and subtracting the longer distance from the shorter one will give us the distance between the buoys
5 years ago
Answered By Vinitkumar p
ans is 6.60.
5 years ago
Answered By Vinitkumar p
suppose the first buoy whose angle of depression is 15° is x meter from light house and another one is Y meter from light house.
tan15= x/30
tan26 = y/30
y-x =6.6 meter
I hope you get the drawing by yourself
5 years ago
Answered By Daunte S
To start this question you should draw a diagram to simplify it and easily determine what needs to be measured. From the diagram below, it is clear that we must find the difference in the length of the bottom side of that triangle based on the angles we are given. The angles given are termed "angles of depression," which means that it is measured starting from looking straight ahead (as if you are on a ledge), and then looking down. Ex: an angle of depression of 15 degrees means that you looked 15 degrees down from eye level.
Now we can begin calculations. The graph shows the relation between the angles to each other. Applying the "Tangent Formula" to both triangles (tanθ = opposite/adjacent), we can find the two side lengths needed. Triangle "a" uses tan(15) = 30/adjacent, rearranged this gives us adjacent = 30/tan(15), which equals 111.96m. Applying the same process to triangle "b" gives us an adjacent length of 61.51m.
So now that we have both lengths, we need to find the difference between which is just a simple subtraction of 111.96 - 61.51 = 50.45m. Since the answer needs to be to the nearest tenth of a meter, the answer is actually 50.5m.
Attached Whiteboard:
Play Drawing5 years ago
Answered By Daunte S
The whiteboard playback seems to be glitched, but examples of this problem can be found online for further clarification. The trick to this question was that the angle of depression cannot be used directly into the formula from the observers point of view which was the error that the person above me made.
For any future trigonometry questions, always try to use a diagram, thinking with shapes makes it much easier.
5 years ago
Answered By Mat B
To answer this question, it helps to know what an angle of depression is. from a horizontal line, (or someone's gaze in a straight line at eye level) an angle of depression opens downward from the horizontal line. So if someone was looking out the lighthouse window straight forward at eye level, it would be a horizontal gaze 30 meters above sea level. If the person looked down towards the buoys, their switch from a horizontal gaze to a downward gaze of 15 or 26 degrees is what creates the angle of depression.
Notice how the light house wall on the left with the window at the top makes 90 degree angles with the eye level of the person in the light house window. It looks like the 30 meter tall light house also forms a right angle with the horizontal lines at sea level that reach out to the buoys. So there are a few right angle triangles to work with here.
The distance we are solving for is between the two buoys down at sea level. Well there are triangles formed by the 30 meter verticle wall of the light house, the line of sight from the light house window down to the buoys, and the horizontal line at sea level going out to the buoys.
So if we use trigonometry some of this can form a solvable equation. We need to work with angles of the triangles formed by the light house the buoys and so on, and we need to use a side length if we can find one. We've got a 3 meter tall window that's part of the triangle so let's try to use that. We don't know the angle from the buoys back to the light house, and from the buoys to the light house window, but what else is there?
How about the angles of depression? There's a right angle formed by the 30 meter vertical wall and the horizontal sight line from the window. Right angles are 90 degrees, and we can find the measure of the angle between the light house wall and the line from the window to the buoys by subtracting 90-15, and 90-26. Those angles will be 75 degrees, and 64 degrees.
Now that we have the measure of one angle and one side, we can start solving for the distances at sea level from the light house to the buoys. Our trig expressions are Soh Cah Toa. Using the angles we just calculated, the 30 meters would be adjacent and the horizontal line at sea level to the buoys would be opposite of the angles.
So Tan75=op/30
to get the "opp" part by itself in the equation, we multiply both sides by 30 and we see
30Tan(75)=opp
plugging that into your calculator gives 111.96 which is the distance to the furthest buoy. We need to subtract the distance of the light house to the furthest buoy (111.96) from that of the light house to the closer buoy to get the distance between the two buoys.
Using the same trig expression can get us closer to the solution. So using the 64 degree angle and the 30 meter measurement we can start manipulating the Tan expression
Tan64= opp/30
and multiply both sides to solve for the opp side
30Tan(64)=opp = 61.51
and subtracting the longer distance from the shorter one will give us the distance between the buoys
111.96-61.51=50.45m
Attached Whiteboard:
Play Drawing