A man 1.5m tall standing on top of a mountain 293.5m high, observes the angle of depression of two flying boats D and C to be 28 degrees and 34 degrees respectively.
Calculate the distance between two boats. correct to 7 decimal places
Correction: Tan28 = h/BD = 295/BD,
BD = 554.81430728.
BC + CD = BD.
437.4 + CD = 554.8,
CD =
By the way my sextant does not work to that accuracy :)
To find the distance between the two boats, we can use the trigonometric relationship involving the angle of depression.
Let's start with boat D. The angle of depression from the man to boat D is 28 degrees. This means that the line of sight from the man's eye to boat D forms a 28-degree angle with the horizontal ground.
Now, we can create a right triangle. The vertical leg of the triangle represents the height of the mountain, which is 293.5m. The diagonal leg represents the distance between the man and boat D. The horizontal leg represents the distance between the man and the base of the mountain.
We can relate these three sides of the triangle using tangent function: tan(angle) = opposite/adjacent.
In this case, the tangent of 28 degrees is the ratio of the height of the mountain to the distance between the man and the base of the mountain. Let's denote the distance between the man and the base of the mountain as 'x'.
tan(28 degrees) = 293.5 / x
We can rearrange this equation to solve for 'x':
x = 293.5 / tan(28 degrees)
Calculating this value, we get:
x ≈ 509.6924106
Next, we move on to boat C. The angle of depression from the man to boat C is 34 degrees. Using a similar approach, we set up the following equation:
tan(34 degrees) = 293.5 / (x + d)
Where 'd' represents the distance between boat D and boat C.
Rearranging this equation gives:
d = (293.5 / tan(34 degrees)) - x
Substituting the previously calculated value of 'x', we can find 'd':
d ≈ (293.5 / tan(34 degrees)) - 509.6924106
Calculating this value, we get:
d ≈ 937.6307442
Therefore, the distance between boat D and boat C is approximately 937.6307442 meters, correct to 7 decimal places.
flying boats??
I think maybe they are at sea level.
looking from 293.5 + 1.5 = 295 above sea level
from base of mountain (point E) straight up to man is 295 meters
man, E , C is right triangle, 90 deg at E
tan 34 = 295/EC
EC = 295 / tan 34
man, E, D same deal
tan 28 = 295/ED
ED = 295 / tan 28
distance CD = ED - EC
to get 7 decimal places carry more than 7 for ED and EC because you lose precision when you subtract.
Make your sketch.I let AB be the height of the mountain,A at the top.I also adde the 1.5 m to 293.5 m to get AB = 295 m
In right angled triangle ACB, sin 34=295/AC
AC = 295/sin34
Angle DCA = 146 and angle DAC = 6 degrees
By the sine law,
DC/sin6=AC/sin 28
You do the button pushing
h1 = 1.5 m
h2 = 293.5 m.
h = h1+h2 = 1.5 + 293.5 = 295 m.
Locate point B at foot of mountain.
Tan34 = h/BC = 295/BC.
BC = 437.4 m.
Tan26 = h/BD = 295/BD.
BD = 605 m.
BC + CD = BD.
437.4 + CD = 605,
CD =