Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 850.
Saturday, February 2, 2013
Problem 850: Triangle, Double side, Angles, 90 Degrees. Special Right triangle, Sides ratio 1:2
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angle B = angle A + 90
ReplyDeleteBy sine law, 2a/sin(A+90) = a/sinA
So, tanA = 1/2
By problem 848, it is approximately 26.5
Great. Now, could you solve this problem without using Trigonometry?
DeleteBx perpendicular to the AB
ReplyDeleteBx and AC meet at point D. Then , angle DBC=angle A , and the triangles ABC ,DBC are similar.
BD/AB=BC/AC=1/2. So ,AB=2AD .Therefore (problem 848) , angle A=26.5 degrees
Image: http://img152.imageshack.us/img152/6397/geogebra5.png
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
ReplyDeletetrapezul isoscel ABCD (problem 848) , angle A=26.5 degrees
Draw line from B, meeting AC in D, with angle(CBD) = 90°.
ReplyDeleteThen angle(ABD) = angle(A), so triangle ABD is isosceles.
AD = x, then BD = x and CD = 2a - x.
Pythagoras in triangle BCD: (2a-x)² = x² + a², giving x = 3/4·a.
BC = a, BD = x = 3/4·a, CD = 2a-x = 5/4·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.) 53,13°.
So angle(A) = 53,13° / 2 = 26,6° (approx.).
Problem 850
ReplyDeleteSuppose CD perpendicular AB (the point D lies on AB).But <ABC=90+<BCD,so <BCD=<A.So
Triangle ADC and triangle CDB is similar, therefore BC/AC=DC/DA=1/2, then <A=26.5
(problem 848).