I take it you mean pole is 5 degrees to the vertical. In that case sketch a diagram and look at the triangle with sides shadow,flagpole and a ray of the sun. You can then use sine rule to get the length of the flagpole.
I take it you mean pole is 5 degrees to the vertical. In that case sketch a diagram and look at the triangle with sides shadow,flagpole and a ray of the sun. You can then use sine rule to get the length of the flagpole.
How in heck can the sun be 15 degrees above the horizon?
Very unclear...either 5 degrees or 15 degrees at top of flagpole...
5 degrees: f = 30SIN(85) / SIN(5)
15 degrees: f = 30SIN(75) / SIN(15)
If you don't "get that", look up the Law of Sines.
AND: any reason for you to call this a "story problem"?
I notice that this term is used more often than Carter has liver pills !
OK; good nuff...
But here's the problem as posted:
"The shadow of a flagpole installed at an angle 5 degrees away from the sun
casts a 30mshadow when the sun is 15 degrees above the horizon."
So we have a right triangle with one of legs = 30.
What are the angles: 5:85 or 15:75 ? Or something else? "5 degrees away" means what?
Well...
If I was the principal of a school where problems that took longer to decipher
than to solve were made up by a teacher...I'd fire that teacher!
It's evident that the student here is learning right triangles associated with the
Law of Sines...why not just give him a triangle with a given angle and a given side...
No, there is NO right triangle. The problem is very clear that one side of the triangle has length 30 m, one angle on that leg 90- 5= 85 degrees and the other 15 degrees. The third angle is, of course, 180- 85- 15= 80 degrees. You can use the sine law to solve for the lengths of the flag pole: x/sin(15)= 30/sin(100).
You call this very clear:
"The shadow of a flagpole installed at an angle 5 degrees away from the sun
casts a 30mshadow when the sun is 15 degrees above the horizon." ?!
The only thing that's clear here is "30m shadow".
If I install a fence post non-vertically, I don't say: my post is x degrees away from the sun...
If anybody thinks it's "normal" for a teacher to make up such a problem when the poor
student is learning about triangles and angles, well then....
Even Biffboy wasn't sure:
"I take it you mean pole is 5 degrees to the vertical. In that case..."