A vertical flagpole stands on a slope that is inclined at an angle of 10 degrees with the horizontal. When the angle of elevation of the sun measures 25 degrees with the horizontal, the shadow of the flagpole down the slope is 40m in length. How tall is the flagpole, to the nearest tenth of a meter?
-
Ok lala--- I've drawn your figure and this is somewhat confusing. Is the flagpole vertical to the slope or to the horizontal? If the flagpole is actually vertical, which we would expect, then there would be an 80 degree difference between the slope elevation and vertical..so you have two right triangle problems to solve...because of supplementary angles...
The shadow is cast on the hypotenuse of the first triangle and, because of || lines you can deduce the angle of elevation to the slope where the flagpole is--is 10 degrees you can get the height from
the relationship slope height/40 m = sin(10). So height where the flagpole sits is 6.946 m above base level.
Now, horizontal distance of the flagpole to observer is= L = 40*cos(10) =39.39m
Now use the tangent function to determine total height.. opp/39.39 = tan(25). Opp = 18.37 m
Now subtract the 6.946m from this answer and you've got your flagpole height = 11.4 m
The shadow is cast on the hypotenuse of the first triangle and, because of || lines you can deduce the angle of elevation to the slope where the flagpole is--is 10 degrees you can get the height from
the relationship slope height/40 m = sin(10). So height where the flagpole sits is 6.946 m above base level.
Now, horizontal distance of the flagpole to observer is= L = 40*cos(10) =39.39m
Now use the tangent function to determine total height.. opp/39.39 = tan(25). Opp = 18.37 m
Now subtract the 6.946m from this answer and you've got your flagpole height = 11.4 m