# 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°

• Your picture suggests that the solution must be based on certain items of information: the two angles shown, and the distance between the two points. If that's what you meant, could you say so explicitly? Commented Oct 18, 2012 at 16:54
• Sorry, forgot that! I've added the details above.
– caw
Commented Oct 18, 2012 at 17:56
• Since the angles involved are not "easy" (like 45° or 60°), you will inevitably need trig functions to compute exactly the height. Hence, without trig functions, either you must do a careful drawing, or you look up in Wikipedia or elsewhere the actual height of Eiffel tower... Commented Jun 15, 2014 at 9:21

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.

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

$$\textrm{Height of building} = \textrm{Distance between finger and thumb} \times \frac{\textrm{Distance from eye to building}}{\textrm{Distance from eye to hand}}$$

• The picture suggests that the answer was to be done by using the two angles and the distance shown. Commented Oct 18, 2012 at 16:58
• Yes, exactly. It must be done with the given information, no thumb available ;)
– caw
Commented Oct 18, 2012 at 17:56

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.

• Thank you very much for giving various explanations!
– caw
Commented Oct 18, 2012 at 19:25
• Another thing is that if you construct C there may be a question of constructing the perpendicular distance from C to line AB, but this is within the realm of grade 5 geometry, or practical tools like a T square as Andre suggested (or a book, or any other right-angled object that is easy to measure).
– zyx
Commented Oct 18, 2012 at 20:17