Thread: Trig. Questions

I need some help on these two problems. I couldnt undestand how to draw a picture for it, but bare with me.

You are watching a fireworks display where you are standing 290 ft behind the launch pad. The launch tubes are aimed away from you at an angle of 65 degrees with the ground. The angle of elevation for you to see the fireworks is 40 degrees.

1.To the nearest ft., what is the horizontal distance from the launch pad to the ignition point of the fireworks?

2. To the nearest ft, what is the height of the fireworks when they ignite?

Code:

                   A(ignite)




B(you)      C(pad) D

BC = 290 ft
angleACD = 65 degrees
angleABC = 40 degrees
AD = vertical distance, CD = horizontal distance

All yours...

Ok but how do i find the horizontal distance. When i did it i got 103 ft.

You're not showing any work...so I don't know what you can do!

Anyway, angle ACB = 180 - 65 = 115; angle BAC = 180 - 40 - 115 = 25.
So you have all angles of triangle ABC plus BC = 290.
Calculate length of AC using Law of Sines.
Then proceed to right triangle ACD.
If you can't handle that, you need classroom help.

Originally Posted by Wilmer

You're not showing any work...so I don't know what you can do!

Anyway, angle ACB = 180 - 65 = 115; angle BAC = 180 - 40 - 115 = 25.
So you have all angles of triangle ABC plus BC = 290.
Calculate length of AC using Law of Sines.
Then proceed to right triangle ACD.
If you can't handle that, you need classroom help.

I get all that but how can you use the sines ratio. its not a right triangle

Look up the Law of Sines!
AC / SIN(40) = 290 / SIN(25)

Originally Posted by Wilmer

Look up the Law of Sines!
AC / SIN(40) = 290 / SIN(25)

Dont think we got that far in our lesson. for AC I got 441

Keerect!
Now with right triangle: vertical = 400, horizontal = 186 (both rounded)
Got that?