Geometry Problem

Click the figure below to see the complete problem 602.

## Monday, May 16, 2011

Subscribe to:
Post Comments (Atom)

skip to main |
skip to sidebar
## Monday, May 16, 2011

###
Geometry Problem 602: Triangle, Angles, 30 and 45 Degrees, Cevian, Congruence

Online Geometry theorems, problems, solutions, and related topics.

Labels:
30 degrees,
45 degrees,
angle,
cevian,
congruence,
triangle

Subscribe to:
Post Comments (Atom)

15

ReplyDeletew.l.o.g. assume AB = DC = 2

ReplyDeleteDraw BN ⊥ AC

Easy to see BN = NC = 1,

Follows DN = DC – NC = 2 – 1 = 1

BN =DN = CN =1 implies

ΔDBC is right angled isosceles and BD = √2

Now apply Sine Rule to ΔADB:

Sin ∠ADB: sin 30° = AB: BD = 2: √2

√2 Sin (30° + x) = 2 sin 30°= 1

Sin (30° + x) = 1/√2 = sin 45°

x = 15°

Rather than an elegant plane geometry solution, if we can use Sine Rule then no construction is really necessary for BC/AB=sin(30)/sin(45)= 1/√2 & BC/CD=sin(45)/sin(105-x). In other words,sin(105-x)=1 since AB=DC or x=15 deg. Isn't that so?

ReplyDeleteVihaan

Referring to my earlier solution:

ReplyDeleteΔDBC is right angled isosceles implies

30° + x = ∠ADB = ∠ABD = 45°, x = 15°

(No need to apply Sine Rule)

Pravin

There's no need to assume AB = CD = 2, we can just put AB = CD = 2a and that makes the solution much valid.

ReplyDeleteLa idea es resolverlo con elementos de euclides y no por trigonometria con teoremas del seno y del coseno...

ReplyDeleteCE=BE=1/2AB where BE is the altitude of Tr ABC. But DE= CD-CE=CE. So <DBE =45 and hence x=15

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka