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A hot air balloon is anchored to the ground by two cables which are 150 feet apart. One cable makes an angle of 62º with the ground and the other cable makes an angle of 38º with the ground. Find the length of each cable (x and y). Show all work clearly and neatly. Round to the nearest tenth.

A hot air balloon is anchored to the ground by two cables which are 150 feet apart-example-1

1 Answer

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The given diagram is

Recall the sine formula for the given triangle is


(150)/(\sin A)=(y)/(\sin B)=(x)/(\sin C)

Substitute the angles as follws.


(150)/(\sin A)=(y)/(\sin62^o)=(x)/(\sin38^o)

By using the triangle sum property, we get


\angle A+\angle B+\angle C=180^o

Substitute known values, we get


\angle A+62^o+38^o=180^o


\angle A=180^o-100^o
\angle A=80^o


(150)/(\sin 80^o)=(y)/(\sin62^o)=(x)/(\sin38^o)


\text{Consider }(150)/(\sin 80^o)=(y)/(\sin62^o)\text{.}
\text{Use the values }\sin 80^o=0.98\text{ and }\sin 62^o=0.88.


\text{ }(150)/(0.98)=(y)/(0.88)


y=\text{ }(150*0.88)/(0.98)
y=134.69\text{ f}eet


\text{Consider }(150)/(\sin 80^o)=(x)/(\sin 38^o)


\text{Use the values }\sin 80^o=0.98\text{ and }\sin 38^o=0.62.
(150)/(0.98)=(x)/(0.62)


x=(150*0.62)/(0.98)
x=94.90\text{ fe}et

Hence the length of x and y is


x=94.9\text{ fe}et


y=134.7\text{ f}eet

A hot air balloon is anchored to the ground by two cables which are 150 feet apart-example-1
User Thomas Kremmel
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