Question 15 Marks
In the given figure, find the angle measure $x.$
Answer
View full question & answer→Sum of all the interior angles of an n-sided polygon $= (n - 2) \times 180^\circ .$
$\text{m}\angle\text{ADC}=180-50=130^\circ$
$\text{m}\angle\text{DAB}=180-115=656^\circ$
$\text{m}\angle\text{BCD}=180-90=90^\circ$
$\text{m}\angle\text{ADC}+\text{m}\angle\text{DAB}+\text{m}\angle\text{BCD}+\text{m}\angle\text{ABC}$
$=(\text{n}-2)\times180^\circ=(4-2)\times180^\circ=2\times186\circ=360^\circ$
$\Rightarrow\text{m}\angle\text{ADC}+\text{m}\angle\text{DAB}+\text{m}\angle\text{BCD}+\text{m}\angle\text{ABC}=360^\circ$
$\Rightarrow130^\circ+65^\circ+90^\circ+\text{m}\angle\text{ABC}=360^\circ$
$\Rightarrow285^\circ+\text{m}\angle\text{ABC}=360^\circ$
$\Rightarrow\text{m}\angle\text{ABC}=75^\circ$
$\Rightarrow\text{m}\angle\text{CBF}=180-75=105^\circ$
$\therefore\text{X}=105$
$\text{m}\angle\text{ADC}=180-50=130^\circ$
$\text{m}\angle\text{DAB}=180-115=656^\circ$
$\text{m}\angle\text{BCD}=180-90=90^\circ$
$\text{m}\angle\text{ADC}+\text{m}\angle\text{DAB}+\text{m}\angle\text{BCD}+\text{m}\angle\text{ABC}$
$=(\text{n}-2)\times180^\circ=(4-2)\times180^\circ=2\times186\circ=360^\circ$
$\Rightarrow\text{m}\angle\text{ADC}+\text{m}\angle\text{DAB}+\text{m}\angle\text{BCD}+\text{m}\angle\text{ABC}=360^\circ$
$\Rightarrow130^\circ+65^\circ+90^\circ+\text{m}\angle\text{ABC}=360^\circ$
$\Rightarrow285^\circ+\text{m}\angle\text{ABC}=360^\circ$
$\Rightarrow\text{m}\angle\text{ABC}=75^\circ$
$\Rightarrow\text{m}\angle\text{CBF}=180-75=105^\circ$
$\therefore\text{X}=105$