Question

In the above figure, ABCD is a rhombus and CDEF is a square. if $\angle$ ABC = 56${}^0$, find:

(i) $\angle$ DAG

(ii) $\angle$ FEG

(iii) $\angle$ GAC

(iv) $\angle$ AGC

Answer 1

Given that, $ABCD$ is a rhombus and $CDEF$ is a square. Also $\angle ABC={56}^o$

To find out: The measure of $\angle DAG,\angle FEG,\angle GACand\angle AGC$

In the rhombus $ABCD$,

$AB=BC=CD=AD$

Also, in square $CDEF$,

$CD=DE=EF=FC$

Hence, $AB=BC=CD=AD=DE=EF=FC$

$DE=AD$

We know that, opposite angles of a parallelogram are equal.

$\therefore \angle ABC=\angle ADC$

Also, $\angle ADE=\angle EDC+\angle ADC$

$\therefore \angle ADE={90}^o+{56}^o={146}^o$

$\left(i\right)In\Delta ADE,$

$DE=AD$

We know that, angles opposite to equal sides of a triangle are equal.

$\therefore \angle DEA=\angle DAE=x\left(Let\right)$

Hence, by interior angle sum property, $\angle ADE+x+x={180}^o$

$\therefore 2x={180}^0-{146}^0$

$\Rightarrow 2x={34}^o$

$\therefore x={17}^o$

Hence, $\angle DAG={17}^o$

$\left(ii\right)\angle FEG=\angle DEF-\angle DEG$

$\Rightarrow {90}^0-{17}^o={73}^o$

Hence, $\angle FEG={73}^o$

$\left(iii\right)$ We know that, adjacent angles of a parallelogram are supplementary.

$\therefore \angle DAB+\angle ABC={180}^o$

$\therefore \angle DAB={180}^o-{56}^o={124}^o$

Also, the diagonal of a rhombus bisects the vertex angle.

$\therefore \angle DAC=\frac{1}{2}\angle DAB$

$\Rightarrow \angle DAC=\frac{1}{2}{124}^o={62}^o$

Now, $\angle GAC=\angle DAC-\angle DAG$

$\therefore \angle GAC={62}^o-{17}^0={45}^o$

Hence, $\angle GAC={45}^o$

$\left(iv\right)In\Delta AGC,$

$\angle AGC+\angle GCA+\angle GAC={180}^o$ [Interior angle sum property]

Also, $\angle GCA=\frac{1}{2}\angle DCB$ [Diagonal of a rhombus bisects the vertex angle]

And $\angle DCB+\angle ABC={180}^o$ [Adjacent angles of a parallelogram are supplementary]

$\therefore \angle GCA=\frac{1}{2}{124}^0={62}^0$

Hence, $\angle AGC+{62}^o+{45}^o={180}^o$

$\therefore \angle AGC={180}^o-{107}^0={73}^o$

Hence, $\angle AGC={73}^o$.