Angle In A Triangle

join BO &angles OCB=OBC=y(OC & OB are radius) similarly angles OED=ODE=x(OE & OD are radius) similarly angles OBE=OEB=30(OE & OB are radius) CBED is a cyclic quardilatral- Therefore, CBE+CDE=180 (cyclic quardilatral property) (y+30)+x=180 x+y=150 we know that BCD=x & ODE=y Therefore , BCD+ODE+CAD=180 x+y+CAD=180 150+CAD=180 (x+y=150,proved above ) CAD=180-150 =30

By drawing a line from B to O, isosceles triangle BOE will be formed. That gives angle BOE = 180-2(30)=120 deg

Angle BOE = Arc BE = 120 deg

Arc CD = 180 deg (semicircle)

The angle formed by two sectors intersecting at the outside of the circle is equal to 1/2 of the difference of the intercepted arcs.

Thus, angle A= 1/2 (Arc CD - Arc BE) = 1/2 (180-120) = 30 deg

Based on the principle. The external angle of a cyclic quadrangle is equal to opposite internal angle.

Angle A= (1/2)(measure of arc DC-measure of arc BE) [Angle formed by two intersecting secants] Angle A= (1/2)(180-120) [arc DC is a semicircle; arc BE=angle EOB] Thus, Angle A=(1/2)(60)=30

Cyclic quadrilateral property: If a side of a cyclic quadrilateral is produced then the angle thus formed is equal to the interior opposite angle .

la parte de arriba no vale jejeje me equivoque =) http://imageshack.com/a/img835/3339/ycfv.jpg

This should be fairly easy this way.

$JoinOB\\ \angle OBE={ 30 }^{ 0 }\quad \Rightarrow \angle BOE={ 120 }^{ 0 }\\ Join\quad CE.\quad Now\quad \angle BCE={ 60 }^{ 0 }\quad (\because \quad \angle BOE={ 120 }^{ 0 })\\ Also,\angle CEA={ 90 }^{ 0 }\quad (\because \angle CED={ 90 }^{ 0 },angle\quad in\quad a\quad semicircle)\\ Now,\quad in\quad \Delta CEA, \quad we \quad have \quad \angle C = 60^0 \quad \angle E = 90^0 \quad \Rightarrow \angle A=30^0$

The correct answer is 30

itz simple maths i just looked at the triangle and circle if the solvers hav notiesd ,there is a chord BE there is a relation between a chord and the angle next to it

it is very easy

Since O is the center of the circle, <AOD becomes 90º (180-90=90º). E is the center of AD, then <EOD = 45º (90-45=45º). <BEO is 30, then <OED = 60º (90-30=60º). <ODE = 75º [180-(60+45)=75º] <ADC = <ACD, therefore <ADC = 75º and <ACD = 75º <CAD = 30º [180-(75+75)=30º]

good question...bit moderate..but solvable

let a angle @ in ODE triangle.......................... consider big triangle...... u will get the answer

First we join O to B. OE = OB as they are radius of a circle.

Angle EBO = Angle OEB =30 ; angle O = 120 by triangle law .

Then <EOD =<BOC ; OE = OD ; <OED = <ODE =75

So <OED + <OEB + <BEA =180 ; <BEA =75

<BEA = <EBA =75 ; therefore <BAE =30

just let angOED=x

=>angODE= x

=>angAEB= (150-x)

=>angCBE= 180-x ..........(BECD is a cyclic quad, sum of opp angles of a cyclic quad is 180 deg)

=>angABE= 180-(angCBE) =180-(180-x) =x

=>angABE + angAEB + angBAE =180

   =>x+(150-x)+angBAE=180

   =>angBAE=180-150=30

step one: triangles EOD and ACD are similar as both have the same angel "D", and share the same sides therefore OE is Parallel to CA Step two: in trapezium BCOE , since OE is parallel to CB therefore angle BCO is 30 and angel EOC is 150 Step three: in trapezium ACOE, since OE is parallel to CA and angle EOC equals 30 therefore angle A equals 30

join EC since triangle COE is an isosceles triangle angle CEO=OCE=15; therefore angle COE=150; COD=180 then angle EOD=30 and COB=30. since triagle EOD is also an isosceles triangle so angle OED=ODE 180-EOD=150; OED and ODE equals 150 divide 2 =75 now we had D as 75 angle COB=30 and this is also an isosceles triangle then OCB = OBC; 180-COB=150 ; 150divide 2=75 meaning C=75 C+D+A=180 75+75+A=180 A=180-150= 30

the total angle of a triangle = 180. and from the figure...it is also 30 degrees...

I got the answer just simply rotating OEB counter-clockwise and voila its the same angle with CAD

add the 2 points B& O. after joining these points i got 5 separate triangles (BOC), (BOE),(DOE),(ABE)&(ACD). Now we can consider Triangle (OED), ODE&OED are same angle, i use it as x,from this triangle we get EOD=180-2x. Now we can consider triangle OBE&OEB =30,so BOE=120, then in triangle BOC, B&C is same,we can marked angle B&C as y , by calculating BOC=2x-120. Now at point o, summation of all angles is equal to 180. BOC+BPE+EOD=180. or x+y=150. then at triangle ACD, A+x+y=180or A=30. So, finally in triangle ACD, angle CAD=180-

30

i like it

let the wanted angle is x ,so concider cBED is cyclic quad so x= 180- the sum of angle B and E to know the other steeps contact at [email protected]

steps:

1. join points BO to form triangle BOE(an isosceles triangle)

2. BO = OE;Angle OBE = OEB = 30 deg

3. similarly CO = OD = BO = EO therefore

4. angles BOC = EOD = BEO = 30 deg; angles CBO = OED = (180 - 30)/2 = 75 deg

5. angles ABE = AEB = 180 - (30+75) = 75 ;therefore

angle BAE = 180 - (75+75) = 30 deg....Check

Draw line BO, angle EBO =30, In quad BECD, 30+CBO+BCD+30+OED+EDC=360, since OED=ODE (angles opposite equal sides of a triangle) similarly BCO=CBO; 2(BCO+CDO)=300 and BCO+CDO= 180-A, 180-A=150 hence A=30.

By drawing a line from B to O - BO is radius and EO aslo radius, Hence <OBE and <OEB are equal i.e <OBE=30. Hence <BOE=120. OC=OB(Radius) so <OCB=<OBC = x OE=OD(Radius) so <ODE=<OED=y Let <COB=k1 and <DOC=k2 Now <COB+<BOE+<DOE=180 ===> k1+k2+120=180===> k1+k2=60 Now x+x+k1=180 and y+y+k2=180 So 2x+2y+k1+k2= 360 ===> 2x+2y=300 ===> x+y=150. Now in Triangle ACD x+y+<A = 180 ===> <A=30

join BO &angles OCB=OBC=y(OC & OB are radius) similarly angles OED=ODE=x(OE & OD are radius) similarly angles OBE=OEB=30(OE & OB are radius) CBED is a cyclic quardilatral- Therefore, CBE+CDE=180 (cyclic quardilatral property) (y+30)+x=180 , x+y=150 we know that BCD=x & ODE=y Therefore , BCD+ODE+CAD=180, x+y+CAD=180 150+CAD=180 (x+y=150,proved above ) CAD=180-150 =30

Let us join BO. Now,angle OBE = angle OEB = 30 .Similarly,angle Ode= angle oed.angle OCB= angle OBC.y+30+x=180 x+y=150.So,angle A = 30 x is angle deo is x and cbo is y. bedc is cyclic so, b+d and e+c is 180

Let angle OED=X, angle ODE=angle OED(base angles of an isosceles triangle) ,CBED is a cyclic Quad. BCD+BED=180, BCD+30+x=180, BCD=150-x, In triangle CAD, 150-x+x+CAD=180, CAD=180-150, CAD=30 Ans.

itz very easy join the CE now let suppose radius of circle be R then CD=2R & CE=R now apply sin rule angle CDE=90 now ....... sinCED/CD=sinCDE/CE SIN90/2R=SINCDE/R sinCDE=1/2 angle CDE=30 degree

Easy just using that all the angles of triangle are 180 and by isoscales rule its directly comes that A = 30

let <acd = X , <adc = y in triangle abe and triangle adc angle X = angle aeb (angle acd + angle deb = 180 , and also angle deb + angle aeb ) similarly , angle adc = Y = angle abe (angle adc + angle cbe = 180 and also angle ceb + angle abe = 180) so, triangle abc is similar to triangle adc also, angle obc=angle ocb = X ( isosceles triangle ) angle ode = angle oed = Y ( isosceles triangle ) also , angle obe = 30 degree , we know , angle X + 30 drgree + angle Y = 180 degree ( opposite angle of a cyclic quadrilateral )
similarly, angle Y + 30 degree + angle X = 180 degree( opposite angle of a cyclic quadrilateral ) so , angle X + angle Y = 150 degree
also , in triangle , ABE , angle ABE + angle AEB + angle BAE = 180 degree X + Y + angle BAE = 180 or, but , X + Y = 150 degree so angle BAE= 30 degree

join BO &angles OCB=OBC=y(OC & OB are radius) similarly angles OED=ODE=x(OE & OD are radius) similarly angles OBE=OEB=30(OE & OB are radius) CBED is a cyclic quardilatral- Therefore, CBE+CDE=180 (cyclic quardilatral property) (y+30)+x=180 x+y=150 we know that BCD=x & ODE=y Therefore , BCD+ODE+CAD=180 x+y+CAD=180 150+CAD=180 (x+y=150,proved above ) CAD=180-150 =30

In the triangle EOD- EO = OD[radii of the. same circle] now this gives»»»» ang.OED = ang.ODE[ang. opp. to eql sides] let »»» ang.OED=(x)=ang.OD---(1) we know... BCDE is a cyclic quadrilateral.... Thus.... ang.BCD+ang.BED=180° ang.BCD+ang. BEO +ang. OED =180 ang. BCD+30+x =180....[by (1)) and given] Now- ang.BCD =180-30-x =150-x---(2) Now- In tri.ACD--- ang.ACD+ang.ADC+ang.CAD=180 Now.. (150-x)+(x)+CAD=180 -x and+x cancelled out to give- CAD +150=180 CAD= 180-150 =30°

angle ODE = angle OED (angles opp. to equal sides) = x angle CBE = 180 - x (opp. angles of cyclc quad.) angle BEA = 150 - x (linear pair with ang. BED = 30 +x ) now, angle BAE = 180 - (x + 150 - x ) = 30 (angle sum prop.)

Join points CE,we find trangle x COE issocellaes trianle where anles OCE &CEO are 15 degree leaving angle COE=150 degree,accordingly angleEOD+30 degree. Again anles ODE and angle OED =75 degree each,From anle BEO which is 30 degree ( given),we get anleBEA 75 degree and finally angle BAE _ 30 degree ANS

K.K.GARG,INDIA

DEO is an equilateral triangel so each angle degrees = 60. angle BEA=180-angle DEO-angle BEO= 180-60-30=90 angle EBA is different line angle of BEO so the degrees is 60 So the angle CAD = angle BAE = 180-angle BEA-angle EBA = 180-90-60= 30