Angles (Sum of Angles)

First

The figure below consists of 3 straight lines. What is the value of  ∠a + ∠b + ∠c?

Method 1

Using Vertically Opposite Angles, the angle in between ∠a & ∠c is equal to ∠b.

Using Adjacent Angles on a Straight Line, ∠a + ∠b + ∠c = 180° because ∠a, ∠b & ∠c lie on a straight line

Method 2

Using Vertically Opposite Angles, the other 3 angles are equal to ∠a + ∠b + ∠c.

Using Angles at a Point:

2 (∠a + ∠b + ∠c) = 360° or 2∠a + 2∠b + 2∠c = 360°

Therefore, ∠a + ∠b + ∠c = 360° / 2 = 180°

Second

The figure below is not drawn to scale. It is made up of 4 identical squares. find the sum of the 3 marked angles.

Label the topmost angle as a, the one directly beneath is as b and the one on its right as c.

b = 45 (angle bisector of a square)
a + c = 90 (double alternate angles)
a + b + c = 45 + 90 = 135


Third
The figure below is not drawn to scale. It shows two triangles inside a square ABCD. AE = CD. Find ∠BED.

Method 1 (Trans-Diagram)

Extend line AE upwards. Cut each of the two new exterior angles created into halves. Each of the exterior angle halves is equal to the interior opposite base angles of its own isosceles triangle.

(360 - 90) / 2 = 135 (angles at point A)

Method 2 (Algebraic)

AED = (180 - DAE) / 2

AEB = (180 - BAE) /2 

DAB =DAE + AEB = 90

AED + AEB 

= (180 - DAE) / 2 + (180 - BAE) /2 

= (360 - DAE - BAE) / 2

= (360 - DAB) / 2

= (360 - 90) /2

= 135