SOLUTION: Can someone help me with this problem please? Thank you. Driving down a straight highway 74 miles per hour, you see a radio tower ahead and to the right. The angle between the r

Algebra ->  Trigonometry-basics -> SOLUTION: Can someone help me with this problem please? Thank you. Driving down a straight highway 74 miles per hour, you see a radio tower ahead and to the right. The angle between the r      Log On


   

Question 1076207: Can someone help me with this problem please?
Thank you.
Driving down a straight highway 74 miles per hour, you see a radio tower ahead and to the right. The angle between the road and your line of sight to the tower is 19 degree. Ninety seconds later, the angle between the road and your line of sight to the tower is 31 degree. At that moment, how far is the tower?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
The first road location, the second road location, and the tower form a triangle. The distance between the first two road locations is 1.85 miles.

The interior angle at first road location, 19 degrees, given.
The INTERIOR angle at second road location will be 180-31=149 degrees.

THe interior angle at the tower, 180-19-149=12 degrees.

"At that moment", distance from the second road location to the tower, t: Use Law Of Sines...
highlight%28sin%2819%29%2Ft=sin%2812%29%2F1.85%29
Solve for t.



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Draw the triangle to see more clearly.

74%28miles%2Fhour%29%2A%281%2F3600%29%28hours%2Fseconds%29%2A90%2Aseconds=1.85%2Amiles

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

Can someone help me with this problem please?
Thank you.
Driving down a straight highway 74 miles per hour, you see a radio tower ahead and to the right. The angle between the road and your line of sight to the tower is 19 degree. Ninety seconds later, the angle between the road and your line of sight to the tower is 31 degree. At that moment, how far is the tower?
Distance traveled from the matrix%281%2C2%2C+19%5Eo-angle-of-elevation%2C+point%29 to the matrix%281%2C2%2C+31%5Eo-angle-of-elevation%2C+point%29 =  

Using this distance and its opposite angle (12%5Eo), we see that the distance from the matrix%281%2C2%2C+31%5Eo-angle-of-elevation%2C+point%29 to the top of the tower is 2.896908 miles

Using this length, and the 31%5Eo angle, we find that the distance from the matrix%281%2C2%2C+31%5Eo-angle-of-elevation%2C+point%29 to the base of the tower is: highlight_green%28matrix%281%2C2%2C+2.483135%2C+miles%29%29