Question

The measures of two angles of a parallelogram are in the ratio 3:2. Find the measure of each of the angles of the parallelogram.

Answer 1

Hint: Assume that the measure of one of the adjacent angles is x, and the measure of the adjacent other angle is y. Use the property that opposite angles of a parallelogram are equal and the sum of angles of a quadrilateral is $360{}^\circ$. Also, use the fact that the ratio of two adjacent angles is 3:2. Hence form two linear equations in two variables and solve the system to find the values of x and y. Hence find the measures of angles of the triangle.

Complete step-by-step answer:

Given: ABCD is a parallelogram. $\angle \text{A:}\angle \text{B::3:2}$
To determine: $\angle \text{A,}\angle \text{B,}\angle \text{C}$ and $\angle \text{D}$
Let $\angle \text{A=}x{}^\circ$ and $\angle \text{B=}y{}^\circ$
Since opposite angles of a parallelogram are equal, we have
$\angle \text{D=}x{}^\circ$ and $\angle \text{C=}y{}^\circ$.
Now we know that the sum of angles of a quadrilateral is equal to $360{}^\circ$.
Using, we get

& \angle \text{A+}\angle \text{B+}\angle \text{C+}\angle \text{D=360}{}^\circ \\\ & \Rightarrow x+y+y+x=360 \\\ & \Rightarrow 2x+2y=360 \\\ & \Rightarrow x+y=180\text{ (i)} \\\ \end{aligned}$$ Also, we have $\angle \text{A:}\angle \text{B::3:2}$ Hence, $\dfrac{x}{y}=\dfrac{3}{2}$ Cross multiplying, we get $2x=3y$ Dividing both sides by 2, we get $x=\dfrac{3y}{2}\text{ (ii)}$ Substituting the value of x from equation (ii) in equation (i), we get $\begin{aligned} & \dfrac{3y}{2}+y=180 \\\ & \Rightarrow \dfrac{5y}{2}=180 \\\ \end{aligned}$ Multiplying both sides by $\dfrac{2}{5}$, we get $\begin{aligned} & \dfrac{5y}{2}\times \dfrac{2}{5}=180\times \dfrac{2}{5} \\\ & \Rightarrow y=72{}^\circ \\\ \end{aligned}$ Substituting the value of y in equation (ii), we get $x=\dfrac{3}{2}\times 72=108{}^\circ $ Hence the angles of the parallelogram are $72{}^\circ ,108{}^\circ ,72{}^\circ $ and $108{}^\circ $. Note: The opposite angles of a parallelogram are equal Proof:  Consider a parallelogram ABCD as shown above. Join AD. Since CD||AB, we have $\angle \text{CDA=}\angle \text{DAB (i)}$ Also since DB||AC, we have $\angle \text{BDA=}\angle \text{DAC (ii)}$ Adding equation (i) and equation (ii), we get $$\begin{aligned} & \angle \text{CDA+}\angle \text{BDA=}\angle \text{DAB+}\angle \text{DAC } \\\ & \angle \text{CDB=}\angle \text{CAB } \\\ \end{aligned}$$ Similarly, $\angle \text{ACD=}\angle \text{ABD}$ Hence proved.