Triangles Angles: Inside and Outside

Because they form triangles, $(a+b+1)+(c+d+3)+(e+f+5)=180+180+180=540.$

Because they form a full circle, $1+2+3+4+5+6=360.$

By Vertical Angles , $1=4,$ $3=6,$ and $5=2.$

Thus, $2(1+3+5)=360,$ so $1+3+5 = 180.$ Applying this result to the first formula, we have $a+b+c+d+e+f=540-(1+3+5) = 540-180=360.$

here (a+b)=angle 2+angle 3 (ext. angle = sum of opp. int angles) similarily (c+d) = angle 4+angle 5 and (e+f)= angle 6 +angle 1

now (a+b+c+d+e+f) = angle(1+2+3+4+5+6) = 360 (angle around a point)

i' did it in very intrusting way, don't know whether it is right or not :) where three lines are intersecting each other it is making center so sum of angles in a circle is 360 degrees and all others angles are also derived from those so sum of all a, b, c, d, e and f angles should be 360 :) ;)

By the picture, it can easily understand why a+b+c+d+e+f = 360 .

just move the top triangle to the bottom between the both triangle ( wich e,f and c,d). after that, the shapes is trapezium. therefore, we can find the answer. a+b+c+d+e+f = 360, because the number of trapezium (totally) is 360

i imagined it was a hexagon then i imagined again that there are 6 triangles so 3/6 to 1/2 then the sum of all angles in a hexagon is 720 then i divided it into 2 then came up with 360

The points ABCDEF form an irregular hexagon and the sum of the internal angles of it is 720 ((6-2)*180). As a+b+c+d+e+f equals half of the sum of the internal angles ....

I let $O$ denote the point of concurrence.

In degree measure, $\angle FOE + \angle EOD + \angle DOC = 180$

$\implies 180-(e+f)+ \angle AOB + 180-(c+d) =180$

$\implies 180-(e+f)+ 180-(a+b) + 180-(c+d) =180$

$\implies 180+180+180-180 = a+b+c+d+e+f$

$\implies a+b+c+d+e+f = 360.$

Simple. We have 6 angles. 6/3 = 2 triangles. 2 triangles, each worth 180 degrees, contain a total of 360 degrees.

New short solution: q + p = c + d (triangle external angle & opposite angle theorems)

Therefore a + b + c + d + e + f = 360 degrees (triangles OAB, OEF each total 180 degrees. O is intersection point) ...... Q.E.D.

Each inner angle is of the form 180-A-B (or the appropriate other angles) , as well as the reflected angle. That in turn gives you 360 = 2 (180-A-B) + 2 (180-C-D) + 2*(180-E-F) Solving that gives you 360 = A + B + C + D + E + F

Using intrinsic nature of angles :

Consider putting an arrow(or a stick with something marked on one end) on $AD$ .(Assume arrowhead is pointing $A$ )

Rotate it to $AB$ about point $A$ . $\angle a$ is 'counted'.

Repeat the steps until all angles are 'counted'.

The arrowhead will be rotate by a 'circle' - pointing $A$ again - counting $360^ \circ$

Let the centre point be O.

From the choices, the answer has to be a unique number (there is no option to select 'None of the above'). Hence the figure has to be TRUE FOR ANY 3 TRIANGLES.

Just arrange three EQUILATERAL triangles AOB, DOC, & EOF such that AD, BE & FC lie along a line.

Hence the sum is 60 x 6 = 360.

each triangle is similar to others so each angel will equal 60 so f+e=a+b = c+d so 1 equal 3 equal 5 and because 5 = 2 3 = 6 and 1 = 4 by meeting the corner so 1=2=3=4=5=6= 60° so a+b+c+d+e+f equal 360 °