What is a better way to solve these triangles?

Question

Someone brought the following problem to me and I have yet to find a satisfactory solution. Could someone help out? Many thanks.

Let the $\triangle ABC$ represent a hill on horizontal ground $BC,$ with a right angle at the peak $A.$ On the peak a vertical tower $AD$ is erected. If the $\angle ACB=35°,\,\angle ABD=52°$ and $BC=66m,$ find: (a) the length of the tower, (b) the angle of elevation of the top of the tower from $C,$ wrt the ground.

One could of course write down equations involving unknowns and try to solve. In fact that's the only thing I could dream up, using the cosine and sine rules to find two equations in two unknowns. However I think that's very farfetched as it leads to a discouraging system, which I can only carry through by enlisting the help of a CAS. However, this was something given to secondary school children as homework, so I expect it to have a nice, compact solution.

What am I missing? Who will kindly point the way? Again, thank you.

Answer 1

What you describe is impossible. $\angle CBD = 107 > 90$ but as $A,D, $ and $C$ are all to the right of $B$, we must have $\angle CBD < 90$.

Alternatively if $AD$ intersects $BC$ at $M$ then $\angle DMB$ is a right triangle. And $\triangle DMB$ is a right triangle with and obtuse angle $\angle MBD$ and $\angle BDM = \angle BDA = -17$.

Answer 2

Okay what does angle A is right angle mean , is angle BAC IS right angle ? If it's then it is very easy ,let hill is on point e somewhere on bc you can calculate height of hill since in right triangle AEC and AEB $$h cot c + h cot (90-c)=66$$ Once you got h , then you can calculate BE and Angle DBC. And height of tower.

But 52° is too much because this how angle DBE will be 52°+(90°-35°)=107° which is not possible is peak is at 90°.