
Trigonometry Problem
Hi, I would appreciate any help with the problem below.
A river flows due east and a tower [cd] stands on its lefthand bank. From a point a upstream and on the same bank of the river, the angle of elevation of the top of the tower is 60 degrees. From point b at right angles across the river from a, the angle of elevation is 45 degrees. If the height of the tower is 36 metres, find the width of the river, correct to the nearest metre.
Textbook answer is 29 metres.
Workings: there is a right triangle acd on the lefthand bank. [ad] is 21 metres which I got from the tan of 60 degrees. [ac] is 42 metres. This is where I am stuck. I have been trying to look for a second right angle triangle where I can use side [ad] and one other angle to calculate the river's width. Also, from my reading of the question I cannot see how the angle of elevation of the top of the tower from b is just 45 degrees. I know that this is an error on my part, but I just can't see it. I have looked at congruent angles for pointers.
Apologies for not attaching a drawing due to the risk of computer viruses. I hope that I am making some sense and I would appreciate any pointers / tips.
Thank you.

Re: Trigonometry Problem
You don't really have to use trig functions these are "special angles. If the angle at point a is 60 degrees, the right triangle formed is half of an equilateral triangle. If the base, the distance from the tower to point a, is "x" (half of one side of the equilateral triangle) then the hypotenuse is 2x (one side of the equilateral triangle) so the height of the is, by the Pythagorean theorem, $\displaystyle \sqrt{(2x)^2 x^2}= x\sqrt{3}$. That is 36 meters so $\displaystyle x\sqrt{3}= 36$, $\displaystyle x= 36/\sqrt{3}= 12\sqrt{3}$. That is, point a is 12\sqrt{3}= 20.7 m or, to the nearest meter, 21 m as you say.
Similarly, 45 degrees is exactly half of 90 degrees a right triangle with one angle 45 degrees has the other acute angle 45 degrees also so is an isosceles triangle. That means that the distance from the tower to b is exactly the same as the height of the tower, 36 m. Now, on the ground, we have a right triangle with one leg the distance from the tower to a, 20.7 m, and hypotenuse of length 36 m. The width of the river is the other leg of that right triangle.

Re: Trigonometry Problem
Hi,
That's a really concise explanation. I especially like your detailed description of the ratio of the sides of triangle acd, which is one half an equilateral triangle. I had not looked at it that way before.
Triangle cab is an isosceles triangle with d as the right angle. That is the key point that I missed, which is a pity as the previous questions I answered in the chapter should have prompted me to interpret it this way.
Thanks once again for taking the time to help.