from the top of a cliff 126m high, the angle of depression of a boat is 20.7deg. how far is the boat from the foot of the cliff?
Let the boat be at a distance of s meter from the foot of the cliff.
Given, 126/s = tan 20.7
=> s = 126/tan 20.7 = 126/0.377 = 334.22 meter.
bobo nung unang sumagot ,
Tony is correct....
:)
YEAH. TONY IS RIGHT :)
To find the distance of the boat from the foot of the cliff, we can use trigonometry. We have the height of the cliff (126m) and the angle of depression (20.7 degrees).
Let's break down the problem into a right triangle. The vertical leg represents the height of the cliff (126m), while the horizontal leg represents the distance from the boat to the foot of the cliff (unknown). The angle of depression (20.7 degrees) is formed by the hypotenuse and the vertical leg.
Now, we can use the tangent ratio, which states that the tangent of an angle is equal to the opposite leg divided by the adjacent leg. In this case, the tangent of the angle of depression is equal to the height of the cliff divided by the distance of the boat from the foot of the cliff.
tan(angle of depression) = opposite/adjacent
tan(20.7 degrees) = 126m/adjacent
To find the distance (adjacent), we rearrange the equation:
adjacent = 126m / tan(20.7 degrees)
Now, let's calculate the distance using the equation:
adjacent ≈ 126m / tan(20.7 degrees)
≈ 126m / 0.364
Using a calculator, we find that the distance of the boat from the foot of the cliff is approximately:
adjacent ≈ 346.1538 meters
Therefore, the boat is approximately 346.1538 meters away from the foot of the cliff.