Heights and Distances — I — problems for RMO and IITJEE Maths

1. A man observes that at a point due south of a certain tower its angle of elevation is 60 degrees. He then walks 100 meters due west on a horizontal plane and finds that the angle of elevation is 30 degrees. Find the height of the tower and his original distance from it.

2. At the foot of a mountain the elevation of its summit is found to be 45 degrees; after ascending 1000 m towards the mountain up a slope of 30 degree inclination, the elevation is found to be 60 degrees. Find the height of the mountain.

3. A square tower stands upon a horizontal plane, from which three of its upper corners are visible, their angular elevations are respectively 45 degrees, 60 degrees and 45 degrees. Show  that the height of the tower is to the breadth of one of its sides as \sqrt{6}(\sqrt{5}+1) to 4.

4. A lighthouse, facing north, sends out a fan-shaped beam of light extending from north-east to north-west. An observer on a steamer, sailing due west, first sees the light when is 5 km, away from the lighthouse and continues to see it for 30\sqrt{2} minutes. What is the speed of the steamer?

5. A man stands at a point X on the bank XY of a river with straight and parallel banks, and observes that the line joining X to a point Z on the opposite bank makes an angle of 30 degrees with XY. He then goes along the bank a distance of 200 meters to Y and finds that the angle ZYX is 60 degrees. Find the breadth of the river.

6. A man, walking due north, observes that the elevation of a balloon, which is due east of him and is sailing toward the north-west, is then 60 degrees; after he has walked 400 meters the balloon is vertically over his head; find its height supposing it to have always remained the same.

The above are some tricky trig problems ! What is the main trick ? I will suggest a very simple hint: draw as good diagrams as possible!

Nalin Pithwa

More Trigonometric Optimization

There is a general rule for solving a system of equations: when you have fewer equations than unknowns, see if you can somehow can get an equation which does the job of several equations. This general principle can, in fact, be applied to the characterization of many other types of triangles. Consider, for example, the following problem:

Question: Suppose ABC is a triangle in which \cos{A}\cos{B}+\sin{A}\sin{B}\sin{C}=1. Prove that

a:b:c=1:1:\sqrt{2} (IITJEE 1986).

Solution: Superficially, the conclusion to be reached involves the sides of the triangle. But, since only their ratios get involved, we can translate it in terms of the angles. We see that a:b:c=1:1:\sqrt{2} is equivalent to saying that

a=b, a^{2}+b^{2}=c^{2}, which, in turn, amounts to saying that the triangle is isosceles and right angled at C. Thus, we have to prove that C=\pi/2 and A=B=\pi/4. (This, by the way, is a helpful habit. See if you can recast the problem, that is, its hypothesis, or its conclusion in various forms. Although mere recasting does not solve the problem, it often inspires you to try certain methods. For instance, now that we have reduced the problem solely in terms of the angles, we can try to solve it by finding the values of \sin{C}, \sin{A}, \cos{A}, etc.)

Now, coming to the problem itself, we are given only one equation (besides the general equation A+B+C=\pi). So somehow we have to cast this equation in a form in which it is equivalent to two separate equations. This can be done by adding and subtracting \sin{A}\sin{B} to get \cos{(A-B)}+\sin{A}\sin{B}(\sin{C}-1)=1 or

(1-\cos{(A-B)})+\sin{A}\sin{B}(1-\sin{C})=0.

The expressions 1-\cos{(A-B)} and 1-\sin{C} are always non-negative. Also, since A, B lie in the interval (0,\pi), \sin{A}, \sin{B} are positive. Therefore the above equation is equivalent to a system of two equations, viz., \cos{(A-B)}=1 and \sin{C}=1. These two equations imply, respectively, that

A=B and C=\pi/2 as desired.

More later,

Nalin Pithwa