Question

Given that $\angle ACD$ is $4^o$ more then, $\angle CDE$ and $\angle CDE$ is ${22}^o$ more then $\angle DAC$. Also, the sum of $\angle BCE$ and $\angle BDE$ is ${26}^o$ and $\angle EBF={80}^o$. Find the value of $x$.

• A. ${40}^o$
• B. ${30}^o$
• C. ${60}^o$
• D. ${75}^o$

Answer 1

Answer: (B)

${30}^o$

In $\Delta$ACD,

$\angle$ACD = $\angle$CDE + $4^{\circ }$

$\angle$CDE = $\angle$DAC + ${22}^{\circ }$

$\Rightarrow$ $\angle$ACD = ${70}^{\circ }$ (Since, sum of all angles in a triangle is ${180}^{\circ }$)

$\angle$CDE = ${66}^{\circ }$

$\angle$DAC = ${44}^{\circ }$ Let $\angle$BCE be a ,then $\angle$BDE is 26-a (given)
Thus, in $\Delta$ CFD,
$\angle$FCD + $\angle$CDF + $\angle$DFC = ${180}^{\circ }$
$\Rightarrow$70 - a + 66 - 26 + a + $\angle$DFC = ${180}^{\circ }$
$\Rightarrow$$\angle$DFC = ${70}^{\circ }$
$\Rightarrow$ $\angle$BFE = ${70}^{\circ }$ (Since, opposite angles are equal)
Now, in $\Delta$ BFE,
$\angle$ EBF + $\angle$ BFE + $\angle$ FEB = ${180}^{\circ }$
$\Rightarrow$ ${80}^{\circ }$ + ${70}^{\circ }$ + x = ${180}^{\circ }$
$\Rightarrow$ x = ${30}^{\circ }$