Do The Angle Chasing Dance

The Angle Chasing Dance

∠FHG =( 180° - ∠FHI ) = ( 180° - 135° ) = 45° So, ∠GFH = ( ∠FGJ - ∠FHG ) = ( 105° - 45° ) = 60° And ∠ DFC = ∠ GFH = 60° So, ∠FCD = ( ∠EDF - ∠DFC ) = ( 119° - 60° ) = 59°

So, ∠ACB = 180° - ∠FCD = 180° - 59° = 121°

So, ∠ABC = 180° - ( ∠ACB + ∠BAC ) = 180° - ( 121° + 40° ) = 19°

So, ∠ABC = 19°.

In the figure, angleJGD and angleEDG are alternate angles. But the difference between the 2 angles is 14 because segment IJ is not parallel to segment EB.

Therefore, we substract 14 from 135 degrees inorder to get angleACB. Which is equal to ( 135 - 14 = 121 )

The sum of all angles of a triangle is always 180 degrees.

Let angleABC = X

                40+121+X=180

                161+X=180

                180-161=X

                 X=19

∠HFG = ∠DFC = 180 - (45 + 75) = 60

∠CAB + ∠ABC = ∠FCD

40 + ∠ABC = 180 - 60 - 61

∠ABC = 19

As <JGF=105, so, <HGF=180-105= 75 Again, as <IHF=135, so, <GHF=180-135= 45 So, <GFH= 180-45-75= 60 Again, <CFD=<GFH= 60 As, <EDF=119, so, <CDF=180-119= 61 So, ACD=<FCD=180-60-61=59 So, <ACB=180-59= 121 So, <ABC=180-121-40= 19 So, answer: <ABC= 19...

All depends on straight line is 180 and the total of the triangle angles is 180. Result is 19.

Easiest question posted on brilliant

Given:- <IHG=135°, <FGJ=105°, <FDE=119°, <BAC=40°

To Find:- <ABC

Solution:-

1) In triangle HGF,

a) <GHF=180°-<IHF=180°-135°=45°     <GHF=45°

b) <HGF=180°-<JGF=180°-105°=75°     <HGF=75°

c) <HFG=180°-(<GHF-<HGF)=180°-(75°-45°)=180°-120°=60°     <HFG=60°

2) <HFG=<DFC {Vertically Opposite Angles} <DFC=60°

3) <FDC=180°-<FDE=180°-119°=61° <FDC=61° {Linear Pair of Angles}

4) In triangle FDC, a) <ACB=<FDC+<CFD=60°+61°=121° <ACB=121° {Sum of two opposite interior angles is equal to the sum of the exterior angle}

5) In triangle ABC, a) <ABC=180°-(<ACB+<BAC)=180°-(121°+40°)=180°-161°=19°

Therefore, the measure of <ABC=19°

FHG = 180 - 135 = 45

FGH = 180 - 105 = 75

HFG = 180 - (45 + 75) = 60

DFC = 60

FCD = 119 - 60 = 59

ACB = 180 - 59 = 121

ABC = 180 - (121 + 40) = 19

in the Figure Angle 180-105=75, 180-135=45, So angle at F is 60 Angle At D is 180-119=61, Angle at ACD is 59 and Angle at ACB=121 angle at B is 19

According to the picture above, what is the measure angle ABC (in degrees) of ? Clarification: are all straight lines. Solution angle HGF = 75, a, angle GHF = 45, angle HFG = 180 – 75 – 45 = 60 Now angle DFC = 60 opposite angles at a F. Consider triangle DFC angle EDF is external angle EDF = angle DFC + angle DCF There fore angle DFC = 119 – 60 = 59
And this is external for triangle ABC therefore angle B = 59 – 40 = 19

Trivial:

FHG is 45, FGH is 75. HFG and DFC are 60, then FDC is 61 and DCF is 59. ACB is 121, and if we know that CAB is 40 then CBA is 180-161=19.

Every single triad of letters is obviously an angle.

Note: I hate solutions written like that, but it's a lazy moment.

Firstly ,angle FHG +180-135=45degree then angle.HGF +180-105=75degree then by Angle Sum Property{ASP} of triangle angleHFG = 180-120=60degree . now angleDFC=angleHFG=60 degree.{vertically opposite angles} angleFDC=180-119=61degree so by ASP again,angleFCD=180-121=59degree now ,angle ACB=180-59=121 degree now to get angle ABC we apply ASP again therefore angleABC=180-40+121=19degree

HENCE......its solved.

GHF-45 HGF-75 HFG-60 GFC-120 DAC-60 ADC-61 ACD-59 ABC-19

Using alternate interior angles, you follow your way to the needed angle.

Using 3 simple equations:

180-((180-135)+(180-105))=60 HFG

180-((180-119)+60)=59 ACD

180-((180-59)+40)=19 ABC

40+121+X=180

            161+X=180

            180-161=X

             X=19

consider in triangle FCD

angle FCD = 40 + x angle DFC = 180 - angle HFG= 180 - ( 180-135) + ( 180 -105 ) = 180 - 45+75 = 180 - 120= 60 angle FDC = 180-119=61

so.... angle DFC + FCD + CDF = 180 60 + 40 + x +61 = 180 x = 19 degree hence done :P

From Given angles in figure,taking adjacent angles calculations as 180- given angles, we get<CDF=61,<FGH =75,<FHG=45 so <GFH =60 gegree,and >CFG =60( opposite angles Now from triangle CDF we get DCF =59 degree and slo angle BCA = 121. Since <BAC =40 ( given) therefore anfgle CBA = 180 - (40 +121)= 19 degree Ans K.K.GARG,India

In the figure, angleJGD and angleEDG are alternate angles. But the difference between the 2

angles is 14 because segment IJ is not parallel to segment EB.

Therefore, we substract 14 from 135 degrees inorder to get angleACB. Which is equal to ( 135 - 14 =121)

The sum of all angles of a triangle is always 180 degrees.

59=40+x

x=19

Find Angle GHF & HGF. by this we can find angle HFG . Angle HFG= Angle DFC. In triangle Def, let angle def = x, then 119+ x + x =180 , x = 30.5 Then Angle FEC + Angle EFC = Angle FCB = 60+30.5+30.5 = 121 In triangle ACB, let let Angle ABC = y. Then 121 + 40 + y =180. y = 19. Hence solved

In triangle HGF

180-135 = 45 < its angle H and 180-105 = 75...... Angles on the straight line

then angle F = 60 sum of angles in the triangle

In triangle CDF : Angle F = 60..... opposite angles.

180-119 = 61 which is Angle D in CDF.... angles on the straight line.

then 60+61 = 121 which is angle C in ABC. ... exterior angle.

180-40-121 = 19 ... Sum of angles in triangle ABC..

40+121+X=180

            161+X=180

            180-161=X

             X=19

In the figure, I first found the angle measures of triangle FGH which are 45, 75, and 60. Angle HFG is 60 degrees and is a vertical angle with angle DAC so angle DAc is equal to 60. next you find the rest of the angle measure for triangle ADC. By doing this you find out that the anlge ACD is eaual to 59 and get that angle ACB is equal to 121. Then you do 180 minus 121 and 40 to get 19 degrees as the measurement for angle ABC

180-135=45,180-105=75,180-(45+75)=60,120-119=61,180-121=59,180-59=121,180-(121+40)=19.

FGH=75,FHG=45=>HFG=60 =>DFC=60(opposite angles)

FDC=61

=>DCF=59

Now,DCF=CAB+ABC =>ABC=19