Geometry Problem

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

Click the figure below to see the complete problem 851.

## Sunday, February 3, 2013

### Problem 851: Triangle, Triple side, Angles, 90 Degrees, Special Right triangle, Sides ratio 1:3

Labels:
90,
angle,
auxiliary line,
ratio,
special right triangle

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Bx 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=1/3. So ,AB=3AD .Therefore (problem 849) , angle A=18.5 degrees

Locate point D on AC such that AB ⊥ BD.

ReplyDeleteTriangle 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

ReplyDeletetrapezul isoscel ABCD (A= 18.5 per problem 849

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 = 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 !!).

ReplyDeleteHenkie 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 secundo

BC/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.

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