Question

In the adjoining figure, it is given that
(i) $AB{\perp }^{r}BF$ and $EF{\perp }^{r}BF$,
(ii) $AC=BC$
(iii) KD is ${\perp }^r$ tp BC and DE.
Find the measure of $\angle$ x.

• A. ${75}^{\circ }$
• B. ${30}^{\circ }$
• C. ${60}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is C ${60}^{\circ }$

Since, $KD\perp BC$ and $DE.$

$\angle CKD=\angle EDK=90°$

$\angle ZBC+\angle BCZ+\angle CZB=180°$

$\angle BCZ=180°-90°-35°=55°$

Now, In $△CKD$

$\angle CKD+\angle BCZ+\angle KDC=180°$

So, $\angle KDC=180°-90°-55°=35°$

$\angle KDE=\angle CDE+\angle CDK$

So, $\angle CDE=\angle KOE+\angle CDK=90°-35°=55°$

In $△FZD,$

$\angle FZD+\angle ZFD+\angle FDZ=180°$

So, $\angle FDZ=180°-90°-25°=65°$

Now, $\angle x+\angle FDZ+\angle EDC=180°$

So, $\angle x=180°-65°-55°=60°$