Geometry Problem

Click the figure below to see the complete problem 504 about Equilateral Triangle, Angles, 45, 60 Degrees.

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Complete Problem 504

Level: High School, SAT Prep, College geometry

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## Sunday, August 15, 2010

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Problem 504: Equilateral Triangle, Angles, 45, 60 Degrees

See also:

Complete Problem 504

Level: High School, SAT Prep, College geometry

## Link List

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

Geometry Problem

Click the figure below to see the complete problem 504 about Equilateral Triangle, Angles, 45, 60 Degrees.

See also:

Complete Problem 504

Level: High School, SAT Prep, College geometry

Labels:
45 degrees,
60 degrees,
angle,
equilateral,
triangle

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in triangle ABD:<A=45,<B=105,<D=30

ReplyDeletewith the sine law

BDsin30=ABsin45;BD=ABsqr(2)

in triangle BDC

with the cosine law

CD²=BC²+BD²-2BC.BD.cos45=AB²

CD=AC

triangle ACD is isoscele

x=15

.-.

let t=AB, it is known that BD=t\sqrt{2} (because tr ABD is 45-30-105).

ReplyDeleteConsecuently CD=t (by tracing CX, altitude of tr BCD and using Pithagorean Theorem).

So tr ACD is isosceles and finally <CDA= < CAD=15.

by mathreyes

draw DG perpendicular to AD, G on extensin of AC

ReplyDeleteABDG cyclic AC = CB = R => C midpoint => AC = CD

arc BG = 2∙A = 120° => arc AB = 60° => C midpoint

ReplyDeleteC origin, A,B and D circular

ReplyDeleteIBCI=ICDI=r

so trianle BCD isosceles and x=15

kadir Altintaş/emirdağ

Draw perpendiculars BP,CQ to AD,BD respectively. Then Tr ABP congruent to Tr BQC SAA. So BP =QC=BQ=QD since < BDP=30. So Tr BQC= Tr DQC. Hence <QDC=45. x=15

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka

Draw perpendiculars BP and CQ to AD and BD respectively. Then these 2 peroenduculars are easily seen to be equal from congruent Tr.s. Also BP=CQ=BQ=1/2 BD (90 60 30 Tr) = QD, so Tr BQC =Tr BQD. Hence<CDQ = 45,'X=15

ReplyDeleteProblem 504

ReplyDeleteIs <ABD=60+45=105 ,so <BDA=30=60/2=(<BCA)/2. Then the point C is the circumcenter of triangle ABD.So AC=CD.Therefore <ADC=<DAC=60-45=15.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

Solution2

ReplyDeleteLet the bisector of < ACB meet AD at E.

Then <BDE <BCE = 30, hence BECD is concyclic.

So x = <CBE = < CAE =15

Sumith Peiris

Moratuwa

Sri Lanka

Solution 3

ReplyDeleteLet X be the foot of the perpendicular from B to AD

If BD = 2p then since <BDX = 30, BX = AX = p and so AB =BC = sqrt2.a

So in Triangle BCD, <B = 45, BC = sqrt2.p and BD = 2p, hence the same is easily seen to be right angled at C.

Hence BDCX is concyclic and x = <XBC = 15

Sumith Peiris

Moratuwa

Sri Lanka