Question

Calculate $\angle AOC,\angle BOD,\angle AOE$ in the figure given below, given that angle $\angle COD={{90}^{\circ }}$, $\angle BOE={{72}^{\circ }}$and AOB is a straight line. \[\]

Answer 1

Hint: We use the fact that the sum of angles subtended by ray on a straight line (called linear pair of angles or supplementary angles if the number of angles is 2 ) is $180{}^\circ$. We take the linear pair of angles $\angle AOE$ and $\angle BOE$ and solve $\angle AOE+\angle BOE={{180}^{\circ }}$ to get $\angle AOE=3x,\angle AOC=x$. Then we take linear triple of angles $\angle AOC+\angle BOD+\angle COD={{180}^{\circ }}$ to get $\angle BOD=y$\[\]

Complete step-by-step answer:
We know from geometry that the couple of angles that lie on a straight line on the same side are known as linear pair or supplementary angles , their sum is $180{}^\circ$. According to the question, line AOB is a straight line. So, the sum of the $\angle AOE$and $\angle BOE$ will be $180{}^\circ $ as they are linear pairs of angles subtended on a straight line by the ray $\overrightarrow{OE}$ . It is also given in the question that $\angle AOE=3x$ and $\angle BOE={{72}^{\circ }}$. So we have
\[\begin{align}
  & \angle AOE+\angle BOE=\angle AOB={{180}^{\circ }} \\
 & \Rightarrow 3x+{{72}^{\circ }}={{180}^{\circ }} \\
 & \Rightarrow 3x={{180}^{\circ }}-{{72}^{\circ }}={{108}^{\circ }} \\
 & \Rightarrow x=\dfrac{{{108}^{\circ }}}{3}={{36}^{\circ }} \\
\end{align}\]
So we obtained the measure of $\angle AOE$ as,
\[\angle AOE=3x=3\times {{36}^{\circ }}={{108}^{\circ }}\]

Now we observe the angles above the line AOB which are $\angle AOC,\angle BOD,\angle BOD.$ We are given in the question that $\angle AOC=x,\angle BOD={{90}^{\circ }},\angle BOD=y.$ These three angles are subtended on the straight line AOB by the rays $\overrightarrow{OC},\overrightarrow{OD}$. So their sum is also ${{180}^{\circ }}$. We have previously obtained $\angle AOC=x={{36}^{\circ }}.$ We use these values and have,
\[\begin{align}
  & \angle AOC+\angle BOD+\angle BOD=\angle AOB \\
 & \Rightarrow x+{{90}^{\circ }}+y={{180}^{\circ }} \\
 & \Rightarrow {{36}^{\circ }}+{{90}^{\circ }}+y={{180}^{\circ }} \\
 & \Rightarrow y={{180}^{\circ }}-{{126}^{\circ }}={{54}^{\circ }}=\angle BOD \\
\end{align}\]
So the measures of the required angle are $\angle AOC={{36}^{\circ }},\angle BOD={{54}^{\circ }},\angle AOE={{108}^{\circ }}$\[\]

Note: We must be careful of the confusion between linear pairs and opposite angles. Linear pairs are the couple of angles that lie on the same side of a straight line say $L$ when another line ${{L}^{'}}$ intersects $L$ whose sum is $180{}^\circ $ . Opposite angles are the couple of angles obtained at the opposite side of the line $L$ and they are equal in measurement.