4.3 Angles in a Quadrilateral
Can you draw a quadrilateral with three angles of 90^\circ and a fourth that is not 90^\circ? Try, and you fail every time. There is a solid reason.
The angle sum is always 360^\circ
Take any quadrilateral PQRS and draw the diagonal PR. It cuts the figure into two triangles, \triangle PQR and \triangle PRS. The angles in each triangle total 180^\circ: \angle 1 + \angle 2 + \angle 3 = 180^\circ, \qquad \angle 4 + \angle 5 + \angle 6 = 180^\circ. Adding, (\angle 1 + \angle 6) + (\angle 3 + \angle 4) + \angle 2 + \angle 5 = 360^\circ. Those four bracketed groups are precisely the four corner angles of the quadrilateral.
The Angle-Sum Property The four angles of any quadrilateral sum to 360^\circ. \angle A + \angle B + \angle C + \angle D = 360^\circ.
This explains the puzzle: if three angles were 90^\circ they would already consume 270^\circ, leaving the fourth no choice but 360 - 270 = 90^\circ. Three right angles compel the fourth.
Worked Example
Three angles of a quadrilateral measure 95^\circ, 110^\circ and 65^\circ. Find the fourth.
The angles add to 360^\circ, so the fourth is 360 - (95 + 110 + 65) = 360 - 270 = 90^\circ.
Worked Example
The four angles of a quadrilateral are in the ratio 2 : 3 : 4 : 6. Find each angle.
Let the angles be 2k, 3k, 4k, 6k. Their sum is 360^\circ, so 2k + 3k + 4k + 6k = 15k = 360 \;\Rightarrow\; k = 24. The angles are 2k = 48^\circ, 3k = 72^\circ, 4k = 96^\circ and 6k = 144^\circ. (Check: 48 + 72 + 96 + 144 = 360^\circ.)