Question

The diagonals of rectangle ABCD meets at O. If angle BOC = 44°. Find angle OAD. ​

Answer 1

SoLuTiOn :

We have,

$\longrightarrow \sf \angle \: BOC + \angle \: BOA = {180}^{o} (Linear \: Pair)$

$\longrightarrow \sf \: {44}^{o} + \angle \: BOA = {180}^{o}$

$\longrightarrow \sf \: \angle \: BOA = \red{{136}^{o} }$

Since diagonals of a rectangle are equal and they bisect each other. Therefore :

$\sf \: In \: \triangle \:AOB \: we \: have :$

OA = OB

$\rightarrow \sf \angle1 = \angle2(Opposite \: Angles)$

$\sf \: In \: \triangle \:OAB \: we \: have :$

$\star \: \sf \angle1 + \angle2 + \angle \: BOA = {180}^{o}$

$\sf \longrightarrow \: 2 \angle1 + {136}^{o} = {180}^{o}$

$\sf \longrightarrow \: 2 \angle1 = {44}^{o}$

$\sf \longrightarrow \: \angle1 = \red{{22}^{o} }$

Since each angle of a rectangle is a right angle.

$\therefore \sf \angle \: BAD = {90}^{o}$

$\longrightarrow \sf \angle1 + \angle3 = {90}^{o}$

$\longrightarrow \sf {22}^{o} + \angle3 = {90}^{o}$

$\longrightarrow \sf \angle3 = \red{{68}^{o} }$

Hence,Angle OAD = 68°

Answer 2

$\huge \boxed{ \fcolorbox{cyan}{red}{Answer : }}$

Given:

$\rm{The \: diagonals \: of \: a \: rectangle \: ABCD \: meet \: at \: O}$

$\sf{also \: < boc = 40°}$

$\sf{∠BOC=∠AOD}$ (vertically opposite triangle)

$\sf{∠AOD=44°}$

$\bf \underline \red{triangle \:AOD}$

$\sf{∠OAD=∠ODA}$(oppoite angle of Isocles triangle)

$\sf{∠AOD+∠OAD+∠ODA=180</strong><strong>°</strong><strong>}$

sum of angles in a traingle is 180°

$\bf{ \huge{ \boxed{ \red{ \tt{substituting \: }}}}}$

$\sf{∠AOD=44°and∠ODA=∠OAD}$

in above equation ⤴

$\sf{44°+∠OAD+∠OAD=180°}</p><p>$

$\sf{∠OAD+∠OAD=180°−44°$

$\sf{2∠OAD=136°}$

$\sf{∠OAD=68°}$

$\sf{ \huge{ \boxed{ \green{ \sf{∠OAD=68° \: }}}}}$