A surveyor standing on top of a cliff found that the angle of depression of a boat as sea is 22°. if the top of the cliff is 15m
1 answer:
Answer:
From the photo above i interpreted the question and drew a triangle to make it easier to solve
Tan is used to solve the question as we were given opposite (as height of the cliff) and to find Adjacent ( the distance between the foot of the cliff to the boat)
Tan θ =Opp/Adj
Tan 22°=15/a
Cross Multiply
aTanθ=15°
Divide both sides by Tan 22
aTan22/Tan22 = 15/Tan 22
a = 15/Tan 22°
a = 37.13 m
Therefore the distance between the foot of the cliff to the boat is 37.13 metres
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Answer:
The coordinates of J are missing. Plus, I don't see options for the conclusions. I made an imaginary point J, which you could correct to form the proper triangle.
Step-by-step explanation:
See the attachment. Plot all the vertices, including a corrected point J and draw the resulting triangle. It might be that AJKL is a slightly smaller, shifted version of AGHI. Enter the correct coordinates and then compare.
