Right Triangles Example
 Question:
 In the figure below, ABC is a right triangle. The angle C is 90^{o}, angle BDC is 45^{o}, angle A is 30^{o}. D is a
point lies on the segment AC and between points A and C. The length of AD is 5, what is the length of BC? (Figure not drawn to scale)
 Solution:
 In the right triangle ABC, since angle A = 30^{o} (Given), angle C = 90^{o} (Given), then angle ABC = 90^{o}  30^{o} = 60^{o}. (In a right triangle, the two acute angles are complementary.)
 Let x = BC, then AB = 2x. (In a right triangle, the length of the side opposite 30^{o} angle is onehalf the length of the hypotenuse.)
 AB^{2} = AC^{2} + BC^{2} (Pythagorean Theorem),
 AC^{2} = AB^{2}  BC^{2} = (2x)^{2}  x^{2} = 2^{2}x^{2}  x^{2} = 4x^{2}  x^{2} = x^{2}(4  1) = 3x^{2}, then AC = sqrt(3)x
 AC = AD + DC, AD = 5 (Given), now we need to find the length of DC.
 In right triangle BDC, angle BDC = 45^{o} (Given), so angle DBC = 90^{o}  45^{o} = 45^{o}.
 Since angle BDC = angle DBC, then BC = DC = x. (In a triangle, if two angles are equal, then the sides opposite these angles are also equal.)
 AC = AD + DC = 5 + DC = 5 + x, AC = sqrt(3)x, then 5 + x = sqrt(3)x, 5 = sqrt(3)x  x = x[sqrt(3)  1],
 x[sqrt(3)  1] = 5, x = 5/[sqrt(3)  1] = 5[sqrt(3) + 1]/{[sqrt(3) 1][sqrt(3) + 1]} = [5 + 5sqrt(3)]/[sqrt(3)^{2}  1^{2}] =
[5 + 5sqrt(3)]/[3  1] = [5+ 5sqrt(3)]/2, so BC = x = [5 + 5sqrt(3)]/2 = (5/2)[1 + sqrt(3)].
 Properties in special right triangle:


 Special product formula:
 (u + v)(u  v) = u^{2}  v^{2}