trigonometry word problem... (please explain)?

Question

You're standing on the 30 ft high roof of the school. You spy Mr. M wlaking by on the ground below 100 ft away. Fortunately, you have your high-powered water cannon with sonar targeting system. At what angle of depression should you fire the cannon in order to soak Mr. M?

so... the answer would be 16.70 degrees, right?

Answer 1

let a = the angle of depression
x = heighth of building = 30 ft
y = distance to Mr. M = 100 ft

if you make a right triangle with right angle being where the building and ground meet, you have the side opposite and the side adjacent, which you can use to set up an equation for tan(gent) a = y / x (opposite/adjacent).

tan a = 100/30
tan a = 10/3
a = inverse tan (10/3)
a = 73.301 degrees

Answer 2

Drink a few quarts of water and you can go after him with your self-contained water cannon.

The angle of depression is arctan(3/10). If you draw a triangle with one leg 100 ft from Mr. M to the edge of the school, and one leg 30 ft high, The angle whose vertex is Mr. M. (and whose hypotenuse is the line of fire) has a tangent of 3/10. Since you are on top of the building, your angle of depression is the same (alt interior angles of a line connecting 2 parallel lines deal).

Answer 3

Failed to check my mail earlier, simply noticed your message =D anyways, I nonetheless stick through my usual answer. The motive was this: The altitude Y of the kite could be calculated the way you suggested, nevertheless it'd be a lot more difficult. The less difficult means would be to first expect ground stage to be the place Dennis is conserving the kite, and then while you in finding the kite's altitude from that floor stage, add the altitude of where Dennis is conserving the kite. Basically, you break up Y into two ingredients: the triangle's leg & an altitude identical to the one who Dennis is protecting the kite at. Sorta rough to give an explanation for... I'm going to supply it once more anyways =D; Let a be the altitude of the place Dennis is preserving the kite. X = 50 ÷ cos30 = 50 ÷ (√three ÷ 2) = (100√three) ÷ 3 Y = (50 &occasions; tan30) + a = 50 × (1 ÷ √three) + a = (50√3) ÷ 3 + a

Answer 4

did not examine my mail before, in basic terms stated your message =D anyhow, I nonetheless stick via my unique answer. the reason replaced into this: The altitude Y of the kite would nicely be calculated the way you stated, even if it may be a lot extra tricky. the easier way must be to first anticipate floor element to be the position Dennis is conserving the kite, and then once you locate the kite's altitude from that floor element, upload the altitude of the position Dennis is conserving the kite. truly, you chop up Y into 2 parts: the triangle's leg & an altitude a twin of the single which Dennis is conserving the kite at. Sorta difficult to describe... i will provide it back anyhow =D; enable a be the altitude of the position Dennis is conserving the kite. X = 50 ÷ cos30 = 50 ÷ (?3 ÷ 2) = (one hundred?3) ÷ 3 Y = (50 × tan30) + a = 50 × (a million ÷ ?3) + a = (50?3) ÷ 3 + a

Answer 5

arctan (30/100) = 16.7°

We have a right triangle formed by the ground, the side of the building (at right angles) and the water cannon stream (the hypotenuse). We know that the (height / distance) = tan (angle) at Mr. M, so take the inverse (arctan) to find the angle.

Answer 6

inverse tangent 30/100=
tan-1 *(3/10)