Altitudes of Obtuse Triangles

Today's question is simple geometry. It deals with triangles, altitudes, sines, and other mathematical functions and objects. The question is quite long, but its answer is not that long. I guess it's another case for the "question theorem" - the length of the question and the length of the answer are inversely proportional.

In triangle ABC, A = 65 degrees, B = 13 degrees, C = 102 degrees. A line perpendicular to AC intersects the line defined by AC at point P. The perpendicular line also passes through point B. The length of PB is 17. Find the area of triangle ABC.

Get your ruler and draw this. If you read correctly, you should draw an obtuse triangle with an altitude outside the triangle. This altitude forms two right triangles: one is outside triangle ABC and one is including triangle ABC.

Since angle BCP is the supplement of angle C, it is 180 - 102 = 78 degrees. Since the outside triangle is a right triangle, we can calculate side BC of triangle ABC:
sin 78 = 17/BC
BC = 17/sin 78 = 17.379

Well, we have one side. To find the area of ABC, we need another side, and then we can use the area formula of 1/2 * a * b * sin C. It never fails.

Now that we have BC, we have two options. Use the law of sines or use a more creative way: find the hypotenuse of triangle APB. I always vote for creativity, so let's find that side.

Angle B is 13 degrees, as given. The angle adjacent to B, CBP, is 12 degrees (the complement of 78 degrees). This makes angle ABP a total of 25 degrees. Now that we have an angle and a side in a right triangle, we can find the other sides. Let's find side AB, because it uses the cosine function, which is... the cosine function:

cos 25 = 17 / AB
AB = 17 / cos 25
AB = 18.757

Now we have everything for the area formula: we have side c (AB), side a (BC), and angle B. Let's plug them all in:
S = 1/2 * c * a * sin B
S = 36.667 sq. units.

Problem solved. Wasn't that hard, was it?

Hope you enjoyed. There are more nice things like this at Super Math Tips.
Nadav

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