Answer:
407 m
Step-by-step explanation:
Refer the attached figure
Karla spots a parked car on the ground at an angle of depression of 30° i.e. ∠BDP = 30°
Now the balloon rises 50 meters.i.e. AB = 50 m
So, the angle of depression to the car is 35 degrees. i.e.∠ADP = 35°
Let DP be the distance of the car prom point P
Let BP be x
In ΔBDP
\(Tan\theta = \frac{Perpendicular}{Base}\)
\(Tan 30^{\circ} = \frac{BP}{DP}\)
\(\frac{1}{\sqrt{3}}= \frac{x}{DP}\)
\(DP= \frac{x}{\frac{1}{\sqrt{3}}}\) --1
In ΔADP
\(Tan\theta = \frac{Perpendicular}{Base}\)
\(Tan 35^{\circ} = \frac{AP}{DP}\)
\(0.70020= \frac{50+x}{DP}\)
\(DP= \frac{50+x}{0.70020}\) --2
So, equating 1 and 2
\(\frac{50+x}{0.70020} = \frac{x}{\frac{1}{\sqrt{3}}}\)
\(\frac{50+x}{x} = \frac{0.70020}{\frac{1}{\sqrt{3}}}\)
\(\frac{50+x}{x} =1.21278\)
\(50+x =1.21278x\)
\(50 =1.21278x-x\)
\(50 =0.21278x\)
\(\frac{50}{0.21278}=x\)
\(\frac{50}{0.21278}=x\)
\(x=234.982\)
Substitute the value of x in 1
\(DP= \frac{234.982}{\frac{1}{\sqrt{3}}}\)
\(DP= 407.0007\)
Hence the car is at a distance of approximately 407 m from point P
