## Saturday, February 2, 2013

### Problem 850: Triangle, Double side, Angles, 90 Degrees. Special Right triangle, Sides ratio 1:2

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 850.

1. angle B = angle A + 90
By sine law, 2a/sin(A+90) = a/sinA
So, tanA = 1/2
By problem 848, it is approximately 26.5

1. Great. Now, could you solve this problem without using Trigonometry?

2. 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=1/2. So ,AB=2AD .Therefore (problem 848) , angle A=26.5 degrees
Image: http://img152.imageshack.us/img152/6397/geogebra5.png

3. 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 (problem 848) , angle A=26.5 degrees

4. 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 = 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.).

5. Problem 850
Suppose 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).