It All Came From Pencils 2

If we draw a line that make a right angle with the red pencil, we have a pentagon. Since the total angle of a N-angled polygon is (n-2). $180^{o}$ , that of a pentagon will be $540^{o}$ .Therefore, the green angle is (540-90-90-40)/2 = $160^{o}$ . The total of green + yellow + blue = $360^{o}$ , so the blue will be $170^{o}$

We have $AB=AC$ , $CE=CD$ and $BF\parallel CG$ , $EH\parallel DI$ .

Suppose $\angle A=a$ and $\angle ECD=b$ .

Connect $BC$ , then $\triangle ABC$ is isosceles, $\angle ACB=90^\circ-\frac{a}{2}$ .

Since $BF\parallel CG$ , so $\angle BCG=90^\circ$ , hence $\angle ACG=180^\circ-\frac{a}{2}$ .

Thus, $y=180^\circ+\frac{a}{2}-b$

Now, substitute $a=40^\circ$ and $b=30^\circ$ you would get $y=170^\circ$ .

As $\triangle$ ABC is isosceles,

$\angle$ ACB = $\dfrac{180-40}{2}$ = $70^\circ$

As $\angle$ BCE is right angle

y = 360 - 90 - 30 - 70 = $170^\circ$

Simple observation is enough to solve this. Sum of internal angles of a triangle equals 180 degrees; Angle BAC is 40, leaving 140 degrees OR 70 degrees for each of the 'base' angles. BCE is 90 degrees. 70 plus 90 and 30 leaves you with 190 degrees. ACD is 170.

if we extend line CG in the direction of the pencil tip A to a point J and bisect angle BAC we see that angle ACJ is 20 and angle JCD is 150 due to the angles on one side of line JG total 180 so angle y (angle ACD) = ACJ +JCD = 170

Triangle ABC is an isosceles triangle, so with 1 angle given, you can already tell all angles of the triangle. Lets make angle s which = 70 deg, because (180-40)/2 = 90 - 20 = 70. Line BC must be perpendicular to Line CE, because if not, the pencil will look bent or the triangle would not be isosceles, which means angle BCE = 90 deg. Angle ECD is given already, so angle ACD must be 90+70+30 = 190 deg. Angle y and angle ACD add up to 360, since they are complimenting each other. We can get angle y by subtracting by ACD. y + 190 deg = 360 deg, so y + 190 - 190 deg = 360 - 190 = 170 = y. there.

You can mace a line from C to B. Since A = 40, the other two angles are equal(since the two sides are equal) making each angle 70. Then, the angle right above it is 90 since line BC is perpendicular to line CG. Then yo would simply perform 360 - (70 + 90 + 30) = 360 - 190 = 170.

Notice the shape of the pencil. If we try to focus on the 2D shape, we see that it is an irregular pentagon. Now we know that the sum of angles of a pentagon is (2 5-4) right angles = (10-4) 90=6*90=540

according to the 2D picture $\angle F = \angle G\ = 90 ^\circ$ . Also it should be obvious that $AB=AC$ , this part of a pencil is always symmetric, that's what we are used to writing with. That makes $\angle ABF$ and $\angle ACG$ equal (Connect $BC$ and think about the figure, you'll notice).

Now let $\angle ABF = \angle ACG = x$ we have $\\x+x+90+90+40=540 \\ 2x=320 \\ x=160$

$\angle ACE + \angle DCE + \angle ACD = 360 \\160+30+y=360\\y+190=360\\y=170$ [solution]

170 because it is less than 180 and more than 160

Here, angleBCE=90 degree. From the triangle ABC, we get, AB=AC. So, angle ABC=angle ACB. So angle ACB=(180-40)/2 degree. angle ACB=70 degree. Now y=360- (70+90+30). So, y=170 degree

The picture to the left was helped because of the right angle formed at the base of the pencil tip.

360 - (70 + 90 + 30) = 170