Grades 11-12 Video Solutions 2011-11&12 Video Solutions 2011 problem27

Video Transcription

Number 27. How many non-congruent isosceles triangles with a base of length 10 are there such that the sine of one angle of this triangle is equal  to the cosine of another angle of this triangle? In my diagram I have a isosceles triangle with a base of 10. I have labeled the congruent angles beta  and the third angle alpha and kind of looks like I have drawn a right isosceles triangle but it doesn't have to be so. The choices upon setting sine of alpha  as one side of the equation is to put cosine of beta on the other side and if I put sine of beta on the left hand side then I can either have a cosine of alpha  or a cosine of beta. So we have three distinct possibilities here and I will go through each case and see if they're possible. So cases, cases 1 and 2 can be  handled at the same time so they can be handled at once. We have the equation here essentially sine of X is equal to cosine of Y and that happens in several  occasions either X is equal to Y and they're both equal to pi over 4 or multiples of pi over 4 but we're in the triangle so we have to limit ourselves  here. We can have X is equal to 2 pi over 3 and Y is equal to pi over 6 or Y is equal to pi over 6 and X is equal to 2 pi over 3 and again we discount the  solutions here that would make the angle bigger than 180 degrees and let's see if these are okay. If X and Y are both equal to pi over 4 then the sum of the angles  here in this triangle will not be equal to pi so in that case 2 beta plus alpha is not equal to pi so this is not okay and if we set one of the angles to be 2  pi over 3 and the other one to be equal to pi over 6 then only in one case with alpha equal to 2 pi over 3 and beta equal to pi over 6 do we have a triangle.  If beta is equal to 2 pi over 3 and alpha is equal to pi over 6 then 2 beta plus alpha is bigger than pi so we don't have a triangle so this is the only  choice here so between 1 and 2 there is one triangle possible and case 3 here is only solved when beta is equal to pi over 4 and it has to be the same angle  that implies that alpha is equal to pi over 2 and we have a right triangle then the right isosceles triangle so that gives us a total of two choices for our  distinct non-congruent isosceles triangles and the answer here would be C.

Video Summary

The problem involves finding non-congruent isosceles triangles with a base of 10 where the sine of one angle equals the cosine of another. By analyzing angle relationships, three cases arise. In the first two cases, setting the angles to specific values, only one valid triangle configuration is possible where alpha equals \(2\pi/3\) and beta equals \(\pi/6\). In the third case, a right isosceles triangle forms when beta equals \(\pi/4\) and alpha equals \(\pi/2\). Therefore, there are only two distinct non-congruent isosceles triangles possible, resulting in the answer.

Keywords

non-congruent isosceles triangles

angle relationships

sine cosine equality

triangle configurations

isosceles triangle cases