Question

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 30o. The balloon rises 50 meters. Now the angle of depression to the car is 35 degrees. How far is the car from point P?

Answer 1

We form 2 similar rt. triangles:

Triangle #1.
X = Hor. side = dist. from P to the car.
Yi = Ver. side.
A = 30o.

Triangle #2.
X = Hor. side = Dist. from P to the car.
Y2 = Y1+50 = Ver. side.
A = 35o.

Use aw of Sines.
sin30/Y1 = sin35/(Y1+50).
Cross multiply:
(Y1+50)sin30 = Y1*sin35
Y1/2+25 = 0.574Y1
Multiply both sides by 2:
Y1+50 = 1.148Y1
1.148Y1-Y1 = 50
0.148Y1 = 50
Y1 = 337.8 m.

tan30 = Y1/X = 337.8/X
X = 337.8/tan30 = 585 m. 714 feet away hope that helps if not tell me.

Answer 2

407 m

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 = (Perpendicular)/(Base)`

`Tan 30^(\circ) = (BP)/(DP)`

`(1)/(√(3))= (x)/(DP)`

`DP= (x)/((1)/(√(3)))` --1

In ΔADP

`Tan\theta = (Perpendicular)/(Base)`

`Tan 35^(\circ) = (AP)/(DP)`

`0.70020= (50+x)/(DP)`

`DP= (50+x)/(0.70020)` --2

So, equating 1 and 2

`(50+x)/(0.70020) = (x)/((1)/(√(3)))`

`(50+x)/(x) = (0.70020)/((1)/(√(3)))`

`(50+x)/(x) =1.21278`

`50+x =1.21278x`

`50 =1.21278x-x`

`50 =0.21278x`

`(50)/(0.21278)=x`

`(50)/(0.21278)=x`

`x=234.982`

Substitute the value of x in 1

`DP= (234.982)/((1)/(√(3)))`

`DP= 407.0007`

Hence the car is at a distance of approximately 407 m from point P