Question

In the adjoining figure, $ABCD$ is a rhombus and $EDC$ is an equilateral triangle. If $\angle DAB={48}^o$ , find
i) $\angle BEC$
ii) $\angle DEB$
iii) $\angle BFC$

Answer 1

R.E.F image

Sol :- Given ABCD is a rhombus and EDC

is an equal $\Delta$ and $\angle DAB={48}^{\circ }$

$\to EDC$ is equal $\Delta$ so

$\angle E=\angle C=\angle D={60}^{\circ }$

$\to \angle A=\angle C={48}^{\circ }$ (in rhombus)

$\to \angle A+\angle D={180}^{\circ }$

$\Rightarrow {48}^{\circ }+\angle D={180}^{\circ }$

$\Rightarrow \angle D=132=\angle B$

Now in $\Delta ECB$ which is isosceles $\Delta$

$\angle C=60+48={108}^{\circ }$

and $\angle E=\angle B=\frac{180-\angle C}{2}$

$\Rightarrow \angle E=\angle B=\frac{180-108}{2}$

$\Rightarrow \angle E=\angle B=36$

Hence $\angle BEC={36}^{\circ }$ --Ans (i)

Now for (ii)$\angle DEB=\angle DEC-\angle BEC$

$=60-36$

$\angle DEB={24}^{\circ }$

for (iii) $\angle BFC$ in $\Delta BFC$

$\angle BFC={180}^{\circ }-\angle ECB-\angle FBC$

$\angle BFC={180}^{\circ }-{48}^{\circ }-36$

$\angle BFC={96}^{\circ }$