Chapter 6.1, Problem 50HP

(a)

To determine

To Find: The $a,b$ and $c$ values if $QRSTVX$ is a regular polygon.

  

Answer to Problem 50HP

  $\angle a={30}^{\circ }$ , $\angle b={90}^{\circ }$ and $\angle c={60}^{\circ }$ .

Explanation of Solution

The sum of interior angles is $(n-2)180$ , n is the no of sides. So for the hexagon it resolves to $4(180)=720$ degrees.

  

We can see that $Q$ is 120 degrees. The sum of all the angles in a triangle is 180 degrees. Therefore in $\Delta XQR$ we have $\angle Q={120}^{\circ }$ . The sides $XQ$ and $QR$ are equal since this is a regular polygon hence, $\Delta XQR$ is isosceles. Therefore, the base angles have to be equal and must the total difference of ${180}^{\circ }-{120}^{\circ }$ must be equal. It is ${60}^{\circ }$ which split two ways and make the angles $a$ and $\angle QRX$ both 30 degrees.

  

  $\angle QRS$ measures ${120}^{\circ }$ as it is an interior angle. $\angle QRX={30}^{\circ }$ we can also find $\angle b$ by subtracting it from ${120}^{\circ }$ . The result will give us $\angle b={90}^{\circ }$ .

  

We see that angle b and angle a are all equal as they are alternate interior angles of the transversal XR and also the segments QR and XS. QR and XS are parallel. This will make the quadrilateral an isosceles trapezoid and as also the base angles are equal.

Angles Q and QRS in the trapezoid are also equal to ${120}^{\circ }$ and the addition of all the angles of a quadrilateral interior angles add up to 360 degree. Base angles RSX and QXS are each ahlf of 360-(120+120) or 60 degrees.

As, RST is interior angle of hexagon given. It is equal to ${120}^{\circ }$ because $\angle RXS$ is ${60}^{\circ }$ and $\angle c={60}^{\circ }$ .

Thus the required angles are $\angle a={30}^{\circ }$ , $\angle b={90}^{\circ }$ and $\angle c={60}^{\circ }$ .

Hence, the required angles are found.