Geometry Problem

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

Click the figure below to see the problem 727 details.

## Friday, February 10, 2012

### Problem 727: Triangle, Isosceles, Congruence, Angles, Measurement

Labels:
angle,
congruence,
isosceles,
measurement,
triangle

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http://img197.imageshack.us/img197/8941/problem727.png

ReplyDeleteLet DD’ is the perpendicular bisector of AC and DD’ cut BC at M

AM is the angle bisector of angle BAC ( see picture)

Draw BK//AC and locate point L on BK such that BL=BA=BD

Note that m(BAM)=m(MAC)=44

Triangle ABL is isosceles => m(BLA)=1/2*(180-92)=44

So m(BAM)=m(BAL)=44 => A, M ,L are collinear

Triangle BKM congruence to LKM ( case ASA)

so BK=KL => triangle BDL is equilateral

and m(DBL)=60 => x=60-44=16

Nice proof, Peter. I applied Ceva's Trigonometric Theorem to Tr. ABC and landed up with: sin(x)*sin(66+x/2)=sin(48-x)*sin(22+x/2)

ReplyDeleteCan anyone tell how this is solved the answer, of course, being x=16 deg.?

Ajit

A trigonometric Proof:

ReplyDeleteB = 48°

∠ABD = 48° - x

ΔABD is isosceles

AD = 2 AB sin(∠ABD/2)= 2 c sin(24° - x/2)

∠BAD = ∠BDA = 66° + x/2

∠DAC = ∠DCA = 22° - x/2

ΔABD is isosceles

b = AC = 2 AD cos (22° - x/2)

b = 4c sin(24° - x/2) cos(22° - x/2)

sin B = 4 sin C sin(24° - x/2) cos(22° - x/2)

sin 48° = 4 sin 44° sin(24° - x/2) cos(22° - x/2)

sin 48° = (2 sin 44°)[2 sin(24° - x/2) cos(22° - x/2)]

sin 48° = (2 sin 44°)[sin(46° - x) + sin 2°]

cos 42° = 2 cos 46° sin(46° - x) + 2 sin 44°sin 2°

cos 42° = 2 cos 46° sin(46° - x) + cos 42° - cos 46°

sin(46° - x) = 1/2

46° - x = 30°

x = 16°

Dear Ajit:

ReplyDeletesin(x)*sin(66+x/2)=sin(48-x)*sin(22+x/2)

=2sin(24-x/2)*cos(24-x/2)*sin(22+x/2)

Observe sin(66+x/2)and cos(24-x/2)get cancelled

since the angles are complementary

We now have

sin(x)= 2sin(24-x/2)*sin(22+x/2)= cos(2-x)-cos(46)

cos(46)=cos(2-x)-sin(x)=sin(88+x)-sin(x)=2cos(44+x)sin(44)

Again cos(46)and sin(44)get cancelled

since the angles are complimentary

So we are left with

1=2cos(44+x)

cos(44+x)= cos(60)

x=16 deg

Pravin

Thanks, Pravin.

ReplyDeleteMy email: ajitathle@gmail.com

Need ur email address.

Ajit

Dear Ajit,

ReplyDeletesin(x)*sin(66+x/2)=sin(48-x)*sin(22+x/2)

sin(x)*cos(24-x/2)=2sin(24-x/2)*cos(24-x/2)*sin(22+x/2), hence cos(24-x/2) get cancelled.

sin(x)=2sin(24-x/2)*sin(22+x/2). Using product to sum formula,

cos(90-x)=cos(2-x)-cos46

cos(2-x)-cos(90-x)=cos46. After using sum to product formula,

-2sin(46-x)*sin(-44)=sin44

-2sin(46-x)*[-sin44]=2sin30*sin44

2sin44*sin(46-x)=2sin30*sin44, hence 2sin44 get cancelled.

Finally sin(46-x)=sin30, then x=16.

Problem 727

ReplyDeleteForms the isosceles trapezoid ABEC (the point Eis the right of point B and above the point C).

(ΒΕ//ΑC,<BAC=88=<ECA). Then AB=BD=EC=BE and BC is bisector <ACE.So ABEC is cyclic. But triangle ABD=triangle CED (AB=BD=CE, AD=DC,<BAD=<ECD).Then BD=DE=BE, so triangle BED is equilateral .Therefore x=<DBC=<DBE-<CBE=60-44=16.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE