Respuesta :
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.
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:
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
[tex]Tan\theta = \frac{Perpendicular}{Base}[/tex]
[tex]Tan 30^{\circ} = \frac{BP}{DP}[/tex]
[tex]\frac{1}{\sqrt{3}}= \frac{x}{DP}[/tex]
[tex]DP= \frac{x}{\frac{1}{\sqrt{3}}}[/tex] --1
In ΔADP
[tex]Tan\theta = \frac{Perpendicular}{Base}[/tex]
[tex]Tan 35^{\circ} = \frac{AP}{DP}[/tex]
[tex]0.70020= \frac{50+x}{DP}[/tex]
[tex]DP= \frac{50+x}{0.70020}[/tex] --2
So, equating 1 and 2
[tex] \frac{50+x}{0.70020} = \frac{x}{\frac{1}{\sqrt{3}}}[/tex]
[tex] \frac{50+x}{x} = \frac{0.70020}{\frac{1}{\sqrt{3}}}[/tex]
[tex] \frac{50+x}{x} =1.21278[/tex]
[tex] 50+x =1.21278x[/tex]
[tex] 50 =1.21278x-x[/tex]
[tex] 50 =0.21278x[/tex]
[tex] \frac{50}{0.21278}=x[/tex]
[tex] \frac{50}{0.21278}=x[/tex]
[tex]x=234.982[/tex]
Substitute the value of x in 1
[tex]DP= \frac{234.982}{\frac{1}{\sqrt{3}}}[/tex]
[tex]DP= 407.0007[/tex]
Hence the car is at a distance of approximately 407 m from point P
