Triangles. math help please?

i need help really bad with this math homework. can someone show me how. my online class doesnt show me how.





A boy is flying a kite and is standing 30 feet from a point directly under the kite. If the string to the kite is 50 feet long, find the angle of the elevation of the kite.


A)126.87° B) 36.87° C)53.13° D)112.6°





The approach pattern to an airport requires pilots to set an 11° angle of descent toward a runway. If a plane is flying at an altitude of 9,500 m, at what distance (measured along the ground) from the airport must the pilot descend?


A)48,873 m B) 9,677.81 m C)49,788 m D) 40, 645 m

Triangles. math help please?
1) hypotenuse = string of kite = 50 ft


ajacent = distance from base of kite = 30 ft





cos x = adj/hypt





x = cos-1 (adj/hypt)


x = 53.13 or c)





2) angle = 11 degrees


altitude = opposite = 9500


runway = adjacent


use tan x = opp/adj


adj = opp/tan 11


adj = 48873 or a)





hope that helps
Reply:1)





Draw the right triangle ABC, where %26lt;B = 90 degrees.





let A represents point of kite.





C represents the boys standing position and BC = 30 ft.





AC = length of kite = 50 ft





angle of elevation = %26lt;ACB





cos(ACB) = BC/AC = 30/50





cos(ACB) = 0.6





angle of elevation,%26lt;ACB = cos^-1(0.6) = 53.13 degrees





2)





In right triangle ABC,





A represents position of plane.





BC = required distance on the ground.





AB = altitude = 9500 m





angle of descent = %26lt;BAC = 11 degrees





cot(BAC) = BC/AB





BC = AB*cot(BAC) = AB/tan(BAC)





BC = 9500/tan(11) = 9500/0.19438 = 48,873 m
Reply:————————————————


Problem A


————————————————


Sketch a triangle and label it to make sense of the problem:





%26lt;|%26gt;


|.\


|...\


|.....\ ← 50ft of string


|.......\ O


|.........\|


|_____/\


....↑30 feet from from the point directly below





angle of elevation is from the ground up:





....\


...θ.\


___(_\





So this tells us:


Angle: from ground, unknown


String: 50 ft hypotenuse


Distance from kite: 30 ft on ground, adjacent to desired angle





to remember what trig function to use, remember:


soh cah toa


(said like "so cuh toe-uh")





meaning:


s = o/h


c = a/h


t = o/a





for


Sin(θ) = Opposite/Hypotenuse


Cos(θ) = Adjacent/Hypotenuse


Tan(θ) = Opposite/Adjacent





In this case, we have the side near the angle (adjacent) and the hypotenuse (the long, angled side), so we use:


cos(θ) = adjacent/hypotenuse





cos(θ) = (distance from kite)/(string length)


cos(θ) = 30/50


cos(θ) = 3/5





The opposite of cos is arccos(x), also labeled cos^-1(x)


arccos(3/5) ≈ 53.1301°





----





Answer:


C) 53.13°





————————————————


Problem B


————————————————


Again, sketch a drawing:





%26gt;=|)%26gt; ---------


..\ ) θ ← angle of descent is from a horizontal line down,


....\ .....|..... not from the ground up like angle of elevation


.....\.....|


.......\...| ← 9500 m above ground


..........\|








We have the side away from the angle this time (opposite)and want the side along ground, which isn't the hypotenuse but the other leg of the triangle. We can set up the triangle in different ways here:


______


|\θ......|


|..\......| ← we can make this triangle from the sky down


|....\....|


|.....\...|


|___θ\ |


↑or one from the ground up





The two places where theta (θ) is at have the same angle, and the top sides both have the same measurement as do the two sides. Depenending on how we set this problem up, we will have to move at least one of these things to an identical spot to set the problem up right. In this case, we want the distance along the ground, but that is the same as the horizontal distance at the of the triangle that was originally drawn for this problem.





Since it gives the angle at the top down and the height (opposite side to that angle) and asks for horizontal distance (adjacent side to that angle), we want opposite and adjacent:





soh cah toa





tangent = o/a is what we want





tan(θ) = opposite/adjacent


tan(angle of descent) = (height above ground)/a


tan(11°) = (9500)/a





a·tan(11°) = [ (9500)/a ]·a


a·tan(11°) = (9500)


a·tan(11°)/tan(11°) = (9500)/tan(11°)


a = (9500)/tan(11°)





a ≈ 48873.263





----





Answer:


A) 48,873 m





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Hope this helps!