Thread: [SOLVED] word problem dealing with angles

Does any1 know how to do these?

Problem 1:
The angle of elevation to the top of a building is found to be from the ground at a distance of feet from the base of the building. Find the height of the building.


Problem 2:
A hot-air balloon is floating above a straight road. To calculate their height above the ground, the balloonists simultaneously measure the angle of depression to two consecutive mileposts on the road on the same side of the balloon. The angles of depression are found to be and .
How high (in feet) is the ballon?

Originally Posted by lsnyder

Problem 1: The angle of elevation to the top of a building is found to be from the ground at a distance of feet from the base of the building. Find the height of the building.

Draw the right triangle. Label the non-right base angle as being eight degrees, and label the base (the "adjacent" side) as being 5000 units.

You are asked for the height (the length of the "opposite") side, so use the tangent ratio.

Originally Posted by lsnyder

Problem 2: A hot-air balloon is floating above a straight road. To calculate their height above the ground, the balloonists simultaneously measure the angle of depression to two consecutive mileposts on the road on the same side of the balloon. The angles of depression are found to be and . How high (in feet) is the ballon?

Draw the nested right triangles. Label the non-right base angles with the given angles. Label the portion of the base between the two non-right base angles as "1" (for "one mile"), and label the rest of the base as "x".

You are asked to find the height. Label this shared side as "h".

You now have two angles and the ratios h/x and h/(x + 1). Use the tangent ratio to solve for h.

If you get stuck, please reply showing how far you have gotten in working through the steps. Thank you!

i missed the hint on my homework.

hint:
Did you convert degrees to radians?

Originally Posted by stapel

Draw the right triangle. Label the non-right base angle as being eight degrees, and label the base (the "adjacent" side) as being 5000 units.

You are asked for the height (the length of the "opposite") side, so use the tangent ratio.


Draw the nested right triangles. Label the non-right base angles with the given angles. Label the portion of the base between the two non-right base angles as "1" (for "one mile"), and label the rest of the base as "x".

You are asked to find the height. Label this shared side as "h".

You now have two angles and the ratios h/x and h/(x + 1). Use the tangent ratio to solve for h.

If you get stuck, please reply showing how far you have gotten in working through the steps. Thank you!

nested right angles?

i got the 1st one,

Problem 1:
702.7

Problem 2, i am lost

For problem 2, the balloon is floating h miles above the ground, and you are asked to solve for h. The first milepost (for which the angle of depression was measured to be 19 degrees) is a distance of x miles from the point on the ground directly beneath the balloon. The second milepost is x + 1 miles away from this point. Hence you have two equations and two unknowns:

Originally Posted by lsnyder

i got the 1st one,

Problem 1:
702.7

Problem 2, i am lost

Hi snyder,


A picture always helps.

okay, that picture definetly helped but i got a question.

if i am finding the depression, do i solve it like it is finding elevation?
what i mean can u sort of switch it if i found it for elevation?
i am not sure if that makes sense.

Originally Posted by lsnyder

okay, that picture definetly helped but i got a question.

if i am finding the depression, do i solve it like it is finding elevation?
what i mean can u sort of switch it if i found it for elevation?
i am not sure if that makes sense.

The angle of elevation = the angle of depression.

These angles are alternate interior angles between the two parallel lines in the figure. And we know that when two parallel lines are cut by a transversal, alternate interior angles are congruent.

Originally Posted by lsnyder

nested right angles?

What picture did you draw, that you didn't get two triangles, one inside the other?

i got the answer now.
anwser to problem 2: 9053


how i got it:

y= ((y/tan 19degrees)+5280) tan 16 degrees
y= 5280 tan 16 degrees/ (1- (tan 16 degrees/tan 19degrees))
y= 9053