A man is measuring the height of trees. His eye level is 6 feet of the ground, but he is standing uphill from the tree he is measuring. He measured an angle of depression from his eye level of 20 degrees and an angle of elevation of 12 degrees. he also has a 200 foot string tied from the base of the tree to a post at eye level from where he is measuring. How tall is the tree?
+128407
The heights of the hill and the post (= the man's eye level ) are irrelevant.
If a 200 foot string is tied is to the base of the tree from the top of the post, this serves as the hypotenuse of a right triangle. And a line drawn from the top of the post to the tree and running parallel to the ground would form one leg of this triangle. Call this distance "D"
And the angle between the string and "D" is 20°
So "D" = 200cos 20 = 187.94 ft
And the partial height of the tree from the base to the point where "D" interesects the tree forms the remaining leg of this right triangle and is given by 200sin 20 = 68.4 ft
And the rest of the tree's height can be found thusly
187.94*tan 12 = 39.95 ft
So.....the total height of the tree is just 68.4 + 39.95 = 108.35 ft
Thanks to Melody for pointing out my small error!!!
+128407
The heights of the hill and the post (= the man's eye level ) are irrelevant.
If a 200 foot string is tied is to the base of the tree from the top of the post, this serves as the hypotenuse of a right triangle. And a line drawn from the top of the post to the tree and running parallel to the ground would form one leg of this triangle. Call this distance "D"
And the angle between the string and "D" is 20°
So "D" = 200cos 20 = 187.94 ft
And the partial height of the tree from the base to the point where "D" interesects the tree forms the remaining leg of this right triangle and is given by 200sin 20 = 68.4 ft
And the rest of the tree's height can be found thusly
187.94*tan 12 = 39.95 ft
So.....the total height of the tree is just 68.4 + 39.95 = 108.35 ft
Thanks to Melody for pointing out my small error!!!
