Trigonometry without sine and cosine Maybe an unusual (and too easy for you) question, but my younger brother is requested to calculate the height of the Eiffel Tower:

Is this possible, given that he has not learned sine and cosine yet (5th grade)?
Details: A-to-B=200m, alpha=65°, beta=41°
 A: It can be done by making a careful drawing. Draw a straight line segment of some convenient length to represent $AB$. Use a protractor and straightedge to draw lines making angles $\alpha$ and $\beta$ as shown. Drop a perpendicular from the meeting point of these two lines to the line through $A$ and $B$. The sensible tool for that is the T-square.  Measure, scale.
If one can assume that the given picture was done to scale, one can even work directly with the picture: Measure $AB$, and the height of the pictured tower, and scale suitably. But it is not safe to rely on the accuracy of a textbook picture. Moreover, it would deprive students of a useful exercise in drawing. 
Remark: If we are concerned that the various tools mentioned above may be used as weapons, we can do the job virtually with geometric software. 
A: You can do it with simple scaling arguments.
Hold your forefinger and thumb a fixed distance apart, and move your hand away from yoru face until the building fits between your finger and thumb. You can then use the following relationship:
$$\frac{\textrm{Height of building}}{\textrm{Distance between finger and thumb}} = 
\frac{\textrm{Distance from eye to building}}{\textrm{Distance from eye to hand}}$$
to calculate that
$$\textrm{Height of building} = \textrm{Distance between finger and thumb} \times 
\frac{\textrm{Distance from eye to building}}{\textrm{Distance from eye to hand}}$$
A: There is a unique triangle $ABC$ with $\angle A = (180 - \alpha)$ and $\angle B=\beta$ and $C$ above the line.  The problem is to construct or calculate $X$ from $A$, $B$, $\alpha$, $\beta$.  For any particular way of giving the angles $\alpha$ and $\beta$ the solution will be relatively easy and does not require trigonometric methods.  For physical drawing it is easier to work with a smaller-scale model of the problem, with smaller value of the distance $AB$ but same angles, and scale up to get the answer.
For a Euclidean construction, $\alpha$ and $\beta$ can be given using a protractor (and the problem is reduced to intersecting two known lines), or as angles somewhere in the plane that can be copied to $AB$ using ruler and compass (and the problem is then the same as the protractor case).
If $\tan \alpha$ and $\tan \beta$ are given or measured, there is a simple algebraic solution.  If the distances to the base of the tower are $a$ and $b$, and height of the tower is $h$, then we are told the ratios $\frac{a}{h}$, $\frac{b}{h}$, and $(b-a)$ and want to know $h$. Subtracting the ratios provides $(b-a)/h$ and hence $h$.  This does not require frightening words like tangent to be used, slopes of lines are enough.
