# On a day with no wind, a hot air balloon hovers at a point abouve a long, straight river. On the west side of the balloon, a sailboat is spotted in the river at an angle of depression of 48 degrees. On the east side, a canoe spots the balloon at an angle of inclination of 29 degrees. The distance between the balloon and the canoe is 650 m. a) What is the height of the balloon? b) What is the distance between the balloon and the sailboat? c) What is the distance between the sailboat and the canoe?

The given question can be represented as a figure as shown in the attachment.

Let h be the height of the balloon above the river.

x be the distance of the balloon from the sailboat and y be the distance between the sail boat and the canoe where y=s+t (from the figure).

Given that the distance between the balloon and the canoe is 650m.

Now we know that,

 sin\theta=\frac{Opposite\ side}{Hypotenuse}

So we can write,

 sin29^0=\frac{h}{650}

i.e.

 h=650sin29^0=650\times 0.4848=315.12\ m

Now similarly,

sin48^0=\frac{h}{x}

i.e.

x=\frac{h}{sin48^0}=\frac{315.12}{0.7431}=424.06\ m

Now we know that,

tan\theta=\frac{Opposite\ side}{Adjacent \ side}

So we can write,

tan48^0=\frac{h}{s}=\frac{315.12}{s}

i.e.

 s=\frac{315.12}{tan48^0}=\frac{315.12}{1.111}= 283.64\ m

And,

tan29^0=\frac{h}{t}=\frac{315.12}{t}

i.e.

 t=\frac{315.12}{tan29^0}=\frac{315.12}{0.5543}=568.50\ m

Finally,

y=s+t=283.64+568.50=852.14\ m

Hence we have the answers as:

a) Height of the balloon = h = 315.12 m

b) Distance between the balloon and the sailboat = x = 424.06 m

c) Distance between the sailboat and the canoe = y = 852.14 m

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