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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 851.
Bx perpendicular to the AB Bx and AC meet at point D. Then , angle DBC=angle A , and the triangles ABC ,DBC are similar.BD/AB=BC/AC=1/3. So ,AB=3AD .Therefore (problem 849) , angle A=18.5 degrees
Locate point D on AC such that AB ⊥ BD.Triangle BDC similar to triangle ABC ( case AA)so BC/AC=BD/AB=1/3 so angle A= 18.5 per problem 849
Construim AD ⊥ AC si luam punctul D astfel incat AD =a , D si B sa fie de o parte si de alta a dreptei AC.Vom obtine trapezul isoscel ABCD (A= 18.5 per problem 849
Draw line from B, meeting AC in D, with angle(CBD) = 90°.Then angle(ABD) = angle(A), so triangle ABD is isosceles.AD = x, then BD = x and CD = 3a - x.Pythagoras in triangle BCD: (3a-x)² = x² + a², giving x = 4/3·a.BC = a, BD = x = 4/3·a, CD = 3a-x = 5/3·a.So triangle BCD is the well-known 3-4-5-triangle!Angle(BDC) = 2·angle(A)From the 3-4-5-triangle the angle is well-known, being (approx.) 36,87°.So angle(A) = 36,87° / 2 = 18,4° (approx.).
Felicitari celor care au raspuns 18.5 (radule, radule !!).Henkie responded best.mes(<A) is half the measure of smallest angle from Egyptian triangle 3-4-5 . Indeed, let ADC be the Egyptian triangle with DC=3, AD=4, AC=5. Draw AB angle bisector of <CAD (B is situated on CD). We have, primo <ABC - <BAC = (<ADB + <BAD) - <BAC ..from exterior angle theorem..= 90 deg,and secundoBC/BD = AC/AD..Angle Bisector Theorem..and from BC/BD = 5/4, BC+BD = 3 (=DC) we get length BC=5/3. So 5= =AC= 3BC.Thus triangle ABC from problem 851is found.Visit too my blog http://ogeometrie-cip.blogspot.ro/ unde am o solutie admirabila pentru problema 1.