1. Karla is riding vertically in a hot air balloon, directly over a point P on the ground. Karla spots a parked car on the ground at an angle of depression of 30^o. The balloon rises 50 meters. Now the angle of depression to the car is 35 degrees. How far is the car from point P?

Dear Student,

Please find below the solution to the asked query:

From given information we form our diagram , As :

Here , Car at point " O "  , so distance between car and point " P " =  OP  =  x m 

We know :  tan $\mathrm{\theta } = \frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$ , So

In triangle POQ we get

$$\begin{gathered} \tan 30°= \frac{\mathrm{PQ}}{\mathrm{OP}} \\ \Rightarrow \frac{1}{\sqrt{3}}= \frac{h}{x} ( \mathrm{We} \mathrm{know} \tan 30°=\frac{1}{\sqrt{3}} ) \\ \mathbf{\Rightarrow }\mathit{h}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{x}}{\sqrt{\mathbf{3}}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{-}\mathbf{-}\mathbf{-}\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{1}\mathbf{ }\mathbf{)} \end{gathered}$$

Now In triangle POR we get 

$$\begin{gathered} \tan 35°= \frac{\mathrm{PR}}{\mathrm{OP}} \\ \Rightarrow 0.7= \frac{h + 50}{x} ( \mathrm{We} \mathrm{know} \tan 35°=0.7 \mathrm{from} \mathrm{table}) \\ \Rightarrow 0.7= \frac{{\displaystyle \frac{x}{\sqrt{3}}} + 50}{x} ( \mathrm{Substitute} \mathrm{value} \mathrm{from} \mathrm{equation} 1 ) \\ \Rightarrow 0.7x = \frac{x}{1.732} + 50 ( \mathrm{We} \mathrm{know} \sqrt{3}=1.732 ) \\ \Rightarrow 0.7x =\frac{x +86.6}{1.732} \\ \Rightarrow 1.2124 x =x +86.6 \\ \Rightarrow 0.2124 x =86.6 \\ \Rightarrow x =\frac{86.6}{0.2124 } \\ \mathbf{\Rightarrow }\mathit{x}\mathbf{ }\mathbf{=}\mathbf{407}\mathbf{.}\mathbf{72} \end{gathered}$$
Therefore,

Distance between car and point " P " = 407.72 m                                        ( Ans )

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