Trigonometry question - anyone ?

Question

A river runs East to west. A tree stands on the edge of one bank and from a point c, the opposite bank and due south of the tree, the angle of elevation of the top of the tree is 28 deg . From point D, 65 m due East of C the angle of elevation of the top of the tree is 20deg. calculate ( A ) The height of the tree, ( B) the width of the river

Answer 1

Nice problem! I had to draw a picture to see what was even going on.

The big mistake everyone else is making is that the point C, D and the tree are not collinear. Point C is directly across the river from the tree, but point D is to the right of Point C! So you have to look at the picture in 3D.

http://homepage.mac.com/fahrenba/yahoo/tree-problem.jpg

The first right triangle gives us this relationship by the pythagorean theorem:

s^2 = r^2 + 65^2

The other two triangles have the following relationships by tangent:

tan(28) = h/r
tan(20) = h/s

There's several things you can do from here. I solved for s in the bottom equation:

s = h/tan(20)

And plugged it into the top equation:

(h/tan(20))^2 = r^2 + 65^2

We can plug in for h now:

tan(28) = h/r
h = rtan(28)

(rtan(28)/tan(20))^2 = r^2 + 65^2

2.134r^2 = r^2 + 4225
1.134r^2 = 4225
r = 61.039 m

So that's the river. To get the height of the tree:

h = r*tan(28)
h = 61.039tan(28) = 32.455 m

Answer 2

Since the height of the tree is the same for each observer, then the tangent equations can be equated:

c*tan28 = d*tan20 thus
d = 1.46 * c

where c is the distance from the tree to C and d is the distance from the tree to D.

Now you can make a triangle with sides 65, c, and 1.46c. The angle between the 65 side and the 1.46c side is then:

invcos x/1.46x = 1/1.46 = 47 degs

Now we determine the sides of the triangle formed by the Tree and C:
The width of the river is 65*tan47 = 70
The height of the tree is 70*tan28 = 37

Answer 3

MRFAHR... seems on target; I didn't review the other answers closely but his answers check:

In the plane of the water his s distance (from tree base to point d) becomes 89.167 m

(sqrt of [{river width}^2 + 65^2}] = sqrt[61.039^2 + 65^2])

arc tan (tree-ht / river-width) = arc tan(32.455 /61.039) =
indeed 28 degrees.

Similarly, arc tan (tree-ht / s) = arc tan (32.455 / 89.169)
= indeed 20 degrees.

(His/her accompanying drawings are quite helpful; the only change I would have suggested is that the lower rt drawing should have pulled "d" further away from the tree to make the angle look smaller than the angle at c.)

Answer 4

height tree is
tan28=h/65
h=65*tan28=35.56m
the width of river
tan20=B/65
B=65*tan20
B=23.65m
good luck

Answer 5

H = height
W = width
Solve the following two equations to get your answers:
H = W tan28
H = (W + 65)tan20