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# Right Triangles Example

Question:
In the figure below, ABC is a right triangle. The angle C is 90o, angle BDC is 45o, angle A is 30o. 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 = 30o (Given), angle C = 90o (Given), then angle ABC = 90o - 30o = 60o. (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 30o angle is one-half the length of the hypotenuse.)
AB2 = AC2 + BC2 (Pythagorean Theorem),
AC2 = AB2 - BC2 = (2x)2 - x2 = 22x2 - x2 = 4x2 - x2 = x2(4 - 1) = 3x2, 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 = 45o (Given), so angle DBC = 90o - 45o = 45o.
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 - 12] = [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) = u2 - v2