cosq = a/h
tanq = o/a
a2 + o2 = h2 (Pythagorean Theorem)
Remember that the equations relating the sides and angles of a right triangle cannot be used for other types of triangles.
cosq = a/h tanq = o/a a2 + o2 = h2 (Pythagorean Theorem) |
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Here is a typical problem involving a non-right triangle: determine the length of the vector C and the angle q that it makes with the x axis.
The most obvious mistake is to write that C2 = A2 + B2, attempting to apply the Pythagorean theorem. Sometimes I see a student attempt this also: suppose the angle between C and the y axis is guessed to be the same as the angle between A and the x axis (30o); they may look similar in size but there is no reason to assume that they are automatically the same. The angle q is then calculated to be 60o. Then the angle between the vectors A and C is calculated to be 30o + 90o + 30o and the student attempts to calculate the length of C by using
When he or she finds that the tangent of 150 degrees is a negative number (-0.58), that should raise a question in his or her mind as to the validity of the equation. Even If the negative sign is ignored, the result is unreasonable in that B is longer than C in the drawing. This attempt to determine the sizes of all the interior angles of the triangle, and then trying to find C and q by stumbling upon the correct information, is a difficult procedure and wastes a lot of time.
It takes practice to be able to "see" the right triangles within a problem, and then to make use of them.
