Proposed Problem

Click the figure below to see the complete problem 302 about Triangle, Angle bisector of 60 degrees.

See also:

Complete Problem 302

Collection of Geometry Problems

Level: High School, SAT Prep, College geometry

## Sunday, June 14, 2009

### Problem 302. Triangle, Angle bisector of 60 degrees.

Labels:
60,
angle bisector,
degree,
triangle

Subscribe to:
Post Comments (Atom)

Tr. ADC and Tr. ADB give us: x/b = sin(C)/sin(B+30) & x/c = sin(B)/sin(C+30). Thus, x/b+x/c= sin(C)/sin(B+30) + sin(B)/sin(C+30) But angles B+30 and C+30 are supplementary. Thus,x/b+x/c= (sin(C)+ sin(B))/sin(B+30). Further, sin(C)=sin(B+60)=sin(B)/2+V3cos(B)/2. So, x/b + x/c = [sin(B)+sin(B)/2+V3cos(B)/2]sin(B+30) which on simplification = V3. Hence, x/b + x/c = V3 or

ReplyDelete1/b + 1/c = V3/x

Ajit Athle: ajitathle@gmail.com

Let BC = a

ReplyDeleteSo,

AD*AD = b*c*{ 1 - [a*a / (b+c)*(b+c)]}

= b*c*{ b*b + 2*b*c + c*c - a*a }/

[(b+c)*(b+c)]

As angle BAC = 60,

a*a = b*b + c*c - b*c

So,

AD*AD

= b*c* {b*b + 2*b*c + c*c - b*b - c*c + b*c}/ [(b+c)*(b+c)]

= b*c*{3*b*c} /[(b+c)*(b+c)]

So,

AD = [Square root of 3]*b*c /[b+c]

So,

x/[Square root of 3] = b*c/[b+c]

So,

[Square root of 3]/x = [b+c] / [b*c]

= [1/b] + [1/c]

Hence the proof.

http://img392.imageshack.us/img392/7320/1001036z.jpg

ReplyDeletedraw DE//AB and DF perpendicular to AC.

ReplyDeletewin DE=AE, DF=x/2

tr DEC and tr ABC are similar (AB//DE), =>DE/c = b-DE/b

=> DE = bc/b+c

from right tr DEF => DE'2 - DE'2/4=x'2/4 => DE=x/v3

sub. x/v3=bc/b+c give the result

p.s. '2 mean exponent 2

v3 mean cube root

Area(ABC)=Area(ABD)+Area(ADC)

ReplyDelete½*b.c.sin(60)=1/2*c.x.sin(30)+1/2*b.x.sin(30)

So b.c.SQRT(3)=b.x+c.x

Divide both side by b.c.x we will get the result

Peter Tran

Altitude BH = sqrt3/2c and altitude DG = x/2. Now use the fact that Tr.s BHC & DGC are similar and that HC = b-c/2 and GC = b- xsqrt3 /2 and the result follows

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka