Lesson A nice Geometry problem solved using Algebra
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<H2>A nice Geometry problem solved using Algebra</H2> <H3>Problem 1</H3>AB and AC are legs of isosceles triangle ABC. Point D lies on AB such that AD = 3, CD = 5, and CB = 5. Find angle A in degrees. <B>Solution</B> <pre> Let x be the side leg of the isosceles triangle ABC: x = AB = AC. Write the cosine law for triangle ABC AB^2 + AC^2 - 2*AB*AC*cos(A) = CD^2, x^2 + x^2 - 2*x*x*cos(A) = 5^2, or 2x^2 - 2x^2*cos(A) = 25, 2x^2*(1-cos(A)) = 25. (1) Next, consider triangle ADC and write the cosine law for it AD^2 + AC^2 - 2*AD*AC*cos(A) = CD^2, 3^2 + x^2 - 2*3*x*cos(A) = 5^2, or x^2 - 6x*cos(A) = 16. (2) From (1), express cos(A) via x 1 - cos(A) = {{{25/(2x^2)}}} ---> cos(A) = 1 - {{{25/(2x^2)}}} = {{{(2x^2-25)/(2x^2)}}}. (3) From (2), express cos(A) via x cos(A) = {{{(x^2-16)/6x}}}. (4) From (3) and (4) {{{(2x^2-25)/(2x^2)}}} = {{{(x^2-16)/(6x)}}}. (5) Since x can not be zero, we can remove the common factor 2x from the denominators in (5). Then, simplifying, we get 3*(2x^2-25) = x*(x^2-16), x^3 - 6x^2 - 16x + 75 = 0. This cubic equation has three real roots: they are x = 3, x = {{{3/2}}} - {{{sqrt(109)/2}}} = -3.72015, x = {{{3/2}}} + {{{sqrt(109)/2}}} = 6.72015. ( see Wolfram Alpha online calculator https://www.wolframalpha.com/input?i2d=true&i=Power%5Bx%2C3%5D+-+6Power%5Bx%2C2%5D+-+16x+%2B+75+%3D+0 ) The negative root does not fit, so we discard it. The root x = 3 also does not fit, so we discard it, too. The only root which works is x = {{{3/2}}} + {{{sqrt(109)/2}}} = 6.72015, approximately. With this x, let's calculate cos(A) using formula (4). It gives cos(A) = 0.72321, approximately. So, A = arccos(0.72321) = 0.76236 radiance, or about 43.68 degrees. <U>ANSWER</U>. Angle A is about 43.68°. </pre> At this point, the solution is complete. 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