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# 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 one-half 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}