Question

In the following figures, find the values of $x,y$ and $z$.

Answer 1

Solution (i):

ABCD is a rectangle. We know that the diagonals of the rectangle are equal and bisect each other.

$⟹BD=AC$ and $AO=BO=OC=OD$

In $△ABO$,

$\angle OAB=\angle OBA={50}^{\circ }$ ---Angles opposite to the equal sides in a triangle are equal.

In $△AOB$,

$\angle AOB+\angle OAB+\angle OBA={180}^{\circ }$ ---sum of interior angles of a triangle

$\therefore \angle AOB=x={180}^{\circ }-{50}^{\circ }-{50}^{\circ }={80}^{\circ }$

Solution (ii):

ABCD is a rectangle. We know that the diagonals of the rectangle are equal and bisect each other.

$⟹BD=AC$ and $AO=BO=OC=OD$

$\angle OCD+\angle DCE={180}^{\circ }$---Angles on a straight line

$\angle ODC={180}^{\circ }-{146}^{\circ }={34}^{\circ }$

In $△DOC$,

$\angle ODC=\angle OCD={34}^{\circ }$ ---Angles opposite to the equal sides in a triangle are equal.

In $△DOC$,

$\angle CDO+\angle DCO+\angle DOC={180}^{\circ }$ ---sum of interior angles of a triangle

$\therefore \angle DOC={180}^{\circ }-{34}^{\circ }-{34}^{\circ }={112}^{\circ }$

$\angle DOC=\angle AOB=x={112}^{\circ }$ ---Vertically Opposite angles

$\angle DCO=\angle OAB=y={34}^{\circ }$ ---Alternate angles

Solution (iii):

ABCD is a Square

$\angle ABC=\angle BCD=\angle CDA=\angle DAB={90}^{\circ }$ ---interior angles of a square are equal to ${90}^{\circ }$

BCE is an equilateral triangle

$\angle EBC=\angle BCE=\angle BEC={60}^{\circ }$

$\angle ECD=\angle BCD+\angle BCE={90}^{\circ }+{60}^{\circ }={150}^{\circ }$

In $△CDE$

$CE=CD$ $⟹\angle CED=\angle CDE$---Angles opposite to the equal sides in a triangle are equal.

$\angle CDE+\angle CED+\angle DCE={180}^{\circ }$ ---sum of interior angles of a triangle

$2\angle CED={180}^{\circ }-{150}^{\circ }={30}^{\circ }$

$\angle CED={15}^{\circ }$

Therefore, $\angle BEC=\angle BED+\angle CED$

${60}^{\circ }=x+{15}^{\circ }$

$x={60}^{\circ }-{15}^{\circ }={45}^{\circ }$

Solution (iii):

BCE is an equilateral triangle

$\angle EBC=\angle BCE=\angle BEC={60}^{\circ }$

ABCD is a Square

$\angle ABC=\angle BCD=\angle CDA=\angle DAB={90}^{\circ }$ ---interior angles of a square are equal to ${90}^{\circ }$

$\angle BCD=\angle ECD+\angle ECB={90}^{\circ }$

$\angle ECD={90}^{\circ }-{60}^{\circ }={30}^{\circ }$

In $△DEC$,

$CE=CD$ $⟹\angle CED=\angle CDE=y$---Angles opposite to the equal sides in a triangle are equal.

$\angle CDE+\angle DCE+\angle DEC={180}^{\circ }$ ---sum of interior angles of a triangle

$\therefore y+y={180}^{\circ }-{30}^{\circ }={150}^{\circ }$

$y={75}^{\circ }$

$\angle ADC=\angle ADE+\angle EDC={90}^{\circ }$

${75}^{\circ }+x={90}^{\circ }$

$x={15}^{\circ }$

$y+z+\angle CEB={360}^{\circ }$ ---Central Angle measures ${360}^{\circ }$

$z={360}^{\circ }-{75}^{\circ }-{60}^{\circ }={225}^{\circ }$

Solution (iv):

$\angle ORS+\angle SRT={180}^{\circ }$---Angles on a straight line

$\angle ORS={180}^{\circ }-{152}^{\circ }={28}^{\circ }$

$y={90}^{\circ }$ ---Diagonals of a Rhombus are orthogonal/perpendicular to each other

In $△ORS$,

$\angle OSR+\angle SOR+\angle ORS={180}^{\circ }$ ---sum of interior angles of a triangle

$\angle OSR={180}^{\circ }-{90}^{\circ }-{28}^{\circ }={62}^{\circ }$

$\angle OSR=\angle OQP=x={62}^{\circ }$ ---Alternate angles

$\angle SRO=\angle OPQ={28}^{\circ }$ ---Alternate angles

$z=\angle OPQ={28}^{\circ }$ ---Diagonal bisects the angle at vertices, in Rhombus