Question

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

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

Answer 1

The correct option is C ${60}^o$

In $△BCZ$,

$\angle CBZ+\angle BCZ+\angle BZC={180}^{\circ }$ ----(sum of interior angles of triangle is ${180}^{\circ }$)

${35}^{\circ }+\angle BCZ+{90}^{\circ }={180}^{\circ }$

$\angle BCZ={180}^{\circ }-{125}^{\circ }={55}^{\circ }$

In $△KCD$,

$\angle DKC+\angle KCD+\angle KDC={180}^{\circ }$ ----(sum of interior angles of triangle is ${180}^{\circ }$)

${90}^{\circ }+\angle KDC+{55}^{\circ }={180}^{\circ }$

$\angle KDC={180}^{\circ }-{145}^{\circ }={35}^{\circ }$

Given, $KD$ is perpendicular to $DE$

$CDE={90}^{\circ }$

$\angle KDC+\angle CDE={90}^{\circ }$

$\angle CDE={90}^{\circ }-{35}^{\circ }={55}^{\circ }$

In $△DZF$,

$\angle DZF+\angle ZDF+\angle DFZ={180}^{\circ }$ ----(sum of interior angles of triangle is ${180}^{\circ }$)

${90}^{\circ }+\angle ZDF+{25}^{\circ }={180}^{\circ }$

$\angle ZDF={180}^{\circ }-{115}^{\circ }={65}^{\circ }$

$CDZ$ is a straight line

From figure $\therefore \angle CDE+\angle EDF+\angle FDZ={180}^{\circ }$

${55}^{\circ }+x^{\circ }+{65}^{\circ }={180}^{\circ }$

$x^{\circ }={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

Option C.