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## Falco276 2 years ago There are many ropes keeping a hot air balloon from floating away before a balloon race. One of these ropes is fixed to the ground at a 45° angle. Another is fixed to the ground at a 30° angle.

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1. Falco276

If the hot air balloon is 21 feet off the ground, what is the distance between the ground directly underneath the balloon and the second rope rounded to the nearest hundredth of a foot? Type the numeric answer only in the box below.

2. Falco276

3. Falco276

@GCR92

4. GCR92

ok let me see

5. Falco276

I think I know. The vertical angle measurement is equal to the horizontal measurement of the right angle.

6. Falco276

21 feet by 21 feet. I just have to find out the measurement to rope 2

7. GCR92

they are asking you the distance between rope 2 and underneath the balloon right?

8. Falco276

yes, exactly.

9. GCR92

just to make sure i understood well, i remind you i'm not english!! hehehe

10. GCR92

ok

11. GCR92

lemme get a paper so i can explain better

12. Falco276

|dw:1358712594988:dw|

13. GCR92

the distance between rope 2 and 1 is x right? as the drawing says

14. Falco276

yes, but I found the half of X which is 45.

15. Falco276

look at the drawing, am I right?

16. Falco276

I only have to fing the measurement that is the question mark.

17. Falco276

*find

18. GCR92

you have 2 angles if you notice

19. GCR92

even though they dont give you a number, you can see that the angle underneath the balloon is 90, a right triangle

20. Falco276

i noticed that.

21. GCR92

apply pitagoras

22. GCR92

ah well you need to find the sides first, nevermind

23. Falco276

sides?

24. GCR92

you gonna solve the problem like this

25. GCR92

x: distance from rope 2 to underneath the balloon - distance from rope 1 to underneath the balloon

26. Falco276

distance from rope 1 to underneath the balloon = 45.

27. Falco276

Im trying to figure out rope 2

28. GCR92

how do you know its 45? 45 is the given angle

29. Falco276

no, not from the angle. On how I got 45 was that the vertical measurement leading to the balloon has the same measurement that leads to rope 1.

30. Falco276

NO NO NO NO!!!!!

31. GCR92

Im solving it, wait

32. Falco276

I have it wrong!

33. GCR92

it has a lot to do with sinus and cosinus

34. Falco276

its 21!!!!!! Its 21!!!!

35. GCR92

well first you need to remember that sinus = op side / h

36. GCR92

we are measuring from rope 1 to balloon now, so sin 30 right?

37. GCR92

i tell you that sin 30 = 1/2

38. GCR92

so you gonna put 1/2 = 21ft / h right?

39. GCR92

solve for h and then you have the hypotenuse

40. GCR92

what do you think you can do next

41. GCR92

there are many ways to solve this problem

42. Falco276

um, okay.

43. GCR92

also remember that all the angles of a triangle = 180, if this doesn't happen, it's not a triangle

44. GCR92

so if you have a 90º angle, a 30º angle...what's the other angle?

45. Falco276

wait, so were trying to find a measurement form rope 1 to the balloon, am I right?

46. GCR92

yep

47. GCR92

i got it already, just waiting to see how you proceed

48. GCR92

that problem is too difficult for you, i mean, they should teach you trigonommetry first, there's no way you can do that without using trigonommetry

49. Falco276

I never took trigonometery before.

50. GCR92

I know you haven't heard of the trigonometric functions

51. GCR92

you need to do it 2 steps

52. Falco276

no, never. I just take basic geometry.

53. GCR92

ok look

54. GCR92

|dw:1358713814089:dw|

55. GCR92

do you remember what was the sinus and what was the cosinus?

56. Falco276

(sigh) I don't remember. (honest, i don't want to be mean to you.)

57. GCR92

hahaha, dont worry about it, i'll explain again, but you should take note of it because you are going to need it in EVERY problem

58. Falco276

okay, I'll take your note.

59. GCR92

|dw:1358714056382:dw|

60. GCR92

Now you need to MEMORIZE this scheme, mathematics are also a subject to study!

61. GCR92

now ill draw you a triangle and you'll tell me which is the oppsite and which is the adjacent side ok?

62. Falco276

ok.

63. GCR92

|dw:1358714498016:dw|

64. Falco276

I think the opposite side is left from angle A the adj side is right from Angle A.

65. GCR92

nope

66. GCR92

just read its name

67. GCR92

opposite = infront of the angle

68. GCR92

adjacent = sidewards the angle

69. GCR92

alongside of it

70. Falco276

fine! Isn't the adjecent side the- How do I Explain it! Isnt the adjecent side the sidewards of the angle?

71. GCR92

draw it

72. Falco276

|dw:1358715165594:dw|

73. GCR92

no, the other one, that's the hypotenuse

74. Falco276

|dw:1358715296994:dw|

75. Falco276

@hba

76. GCR92

yep

77. hba

Yes @Falco276

78. Falco276

@hba please help me!

79. hba

@GCR92 is helping you,Just follow him :)

80. GCR92

I have to explain the whole trigonometry before you are able to solve that problem...

81. Falco276

(sigh) ok.

82. GCR92

I explain as simple as it is, but not simpler

83. Falco276

ok

84. GCR92

ok i put the turbo now

85. GCR92

we are going to solve 2 triangles

86. GCR92

|dw:1358715724015:dw|

87. Falco276

wait. I think I have it.

88. GCR92

we are going to use the sinus to solve this one, why sinus? because the problem has given us the opposite side, wich is 21 ft

89. Falco276

For the 45° triangle

90. GCR92

and sinus = op side / h

91. GCR92

and we can make a relationship

92. GCR92

an equation

93. Falco276

tan Ө = opposite / adjacent tan Ө = height of the balloon off the ground / x tan 45° = 15 / x x = 15 / tan 45° x = 15 / 1 x = 15

94. Falco276

is this correct?

95. GCR92

where did you get 15?

96. Falco276

oh, no i meant 21 (facepalm)

97. GCR92

You could've done it if you had the value of the x side

98. GCR92

but that's what we are looking for

99. GCR92

to use the tangent you need to have the value of 2 sides

100. GCR92

I'd like to continue now

101. Falco276

|dw:1358716138389:dw|

102. GCR92

why?

103. Falco276

|dw:1358716271703:dw|

104. Falco276

|dw:1358716355941:dw|

105. Falco276

now we take the other triangle.

106. GCR92

the adjacent side is not 21, 21 is the only data they give us, you can't invent it

107. GCR92

listen i dont want to be mean either but i want to help you and i want to finish, if you think you don't need to do what i tell you to solve it, go ahead

108. GCR92

sin 30 = 1/2, we do that in order to calculate H

109. GCR92

1/2 = 21 / h

110. GCR92

h= 42

111. GCR92

now we take the cosinus in order to calculate X

112. GCR92

cos 30 = sqrt 3 / 2

113. GCR92

sqrt 3 / 2 = 42 / x

114. GCR92

x = 36.373

115. Falco276

|dw:1358716402610:dw|

116. Falco276

|dw:1358717381175:dw|

117. Falco276

|dw:1358717481086:dw|

118. GCR92

you can never write tan and let it alone

119. GCR92

it is always tan OF SOMETHING

120. GCR92

tan of x, tan of alfa, tan of 20

121. Falco276

so the answer is 21 sq. root 3

122. GCR92

ill solve it

123. GCR92

2 triangles

124. GCR92

take the 30º triangle and then, sin 30 = 1/2

125. GCR92

1/2 = 21 / h, h = 42

126. GCR92

now we have the h of the first triangle

127. GCR92

now we can calculate the distance from the rope to underneath the balloon, that distance will be given by the cosinus

128. GCR92

cos 30 = sqrt 3 / 2

129. GCR92

sqrt 3 /2 = x/42

130. GCR92

x = 36,373

131. GCR92

now we go to the second triangle and apply exactly the same steps but with 45º

132. Falco276

maybe my way of solving this problem was wrong. sorry.

133. Falco276

i tried my best.

134. GCR92

I know but mathematics doesn't magically happen, you need to study a lot( and suffer lol) dont worry

135. GCR92

if you do the same with the second triangle

136. GCR92

sin 45 = sqrt 2 / 2

137. GCR92

sqrt 2 / 2 = 21 / h

138. GCR92

h = 29.698

139. GCR92

now we calculate the cosinus

140. GCR92

cos 45 = sqrt 2 / 2

141. GCR92

sqrt 2 / 2 = x / 28,698

142. GCR92

x = 20,292

143. GCR92

distance from underneath the ballon to the second rope = 36, 373 + 20,292 = 56,665 feet

144. GCR92

and it's solved

145. Falco276

wait, but earlier you said that x is 36- never mind :P

146. GCR92

that's the distance between underneath the balloon and rope 2

147. Falco276

ah, i see.

148. GCR92

and 20,292 ft is the distance between underneath the balloon and rope 1

149. GCR92

sorry

150. Falco276

got it.

151. GCR92

36ft is the distance between rope 2 and 1

152. GCR92

then you calculate from rope 1 to underneath the ballon and it gives you 20 ft

153. GCR92

you add it and you obtain the final result

154. Falco276

okay.

155. GCR92

what i did is totally useless, i mean you can't learn in 20 min, i always believed that a teacher is need in order to learn something, anyway, wish you luck with this!

156. GCR92

needed*

157. GCR92

you did well! at least you had ideas! you only need to learn the theory well

158. GCR92

i also think your teachers are crazy though

159. Falco276

XD Thanks Gonzalo! I promise to keep the theory at heart. Yes, I finally understood the answer and also the work. I'll tell you if need help on anything else.

160. GCR92

You are welcome, sure, see you! :D

161. Falco276

thanks! (thumbs up)

162. GCR92

I also can be wrong, i mean if your teachers yells at you it is my fault, just don't take that in account, it's been ages since i dont take trigonometry :p

163. Falco276

oh, i understand, dont worry! my teachers are really nice.

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