Question
$ABCD$ is a trapezium such that $AB\|CD$, $\angle\text{D},\angle\text{A}:\angle\text{D}=2:1,\angle\text{B}:\angle\text{C}=7:5$ Find the angles of the trapezium.
✓
Answer
Let $ABCD$ is a trapezium, where $AB \| CD$
Let the angles $A$ and $D$ be of measures $2x$ and $x$, respectively
Then, $2x + x = 180^\circ$
$\Rightarrow 3x = 180^\circ$
$\Rightarrow\text{x}=\frac{180^\circ}{3}=60^\circ$
$\Rightarrow\angle\text{A}=2\times60^\circ=120^\circ,\angle\text{D}=60^\circ$
Again, let the angles $B$ and $C$ be $7x$ and $5x$ respectively
Then, $7x + 5x = 180^\circ$
$\Rightarrow 12x = 180^\circ$
$\Rightarrow x = 15$
Thus, $\angle\text{B}=7\times15=105^\circ$
And $\angle\text{C}=5\times5=75^\circ$
Let the angles $A$ and $D$ be of measures $2x$ and $x$, respectively
Then, $2x + x = 180^\circ$
$\Rightarrow 3x = 180^\circ$
$\Rightarrow\text{x}=\frac{180^\circ}{3}=60^\circ$
$\Rightarrow\angle\text{A}=2\times60^\circ=120^\circ,\angle\text{D}=60^\circ$
Again, let the angles $B$ and $C$ be $7x$ and $5x$ respectively
Then, $7x + 5x = 180^\circ$
$\Rightarrow 12x = 180^\circ$
$\Rightarrow x = 15$
Thus, $\angle\text{B}=7\times15=105^\circ$
And $\angle\text{C}=5\times5=75^\circ$
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