Question

please solve this question.​

Answer 1

$\huge \star \: \underline{ \red{ \sf \: solution : }}$

⊙ Given :

• Rectangle
• Angle BOC = 70°
• Angle ODC

⊙ To Find :

• Angle OAB

⊙ Solving :

$\\ \large \tt \mapsto \: \angle doc + 70 \degree = 180 \degree \\ (linear \: \: pair) \\ \\ \large \implies \tt \: \angle doc = 180 \degree - 70 \degree \\ \\ \large \sf \therefore \: \: \angle doc = 110 \degree$

Since, it is a rectangle

• Triangle DOC = AOB
• Triangle DOA = COB

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$\\ \large \leadsto \sf \: in \: \triangle doc :$

• Angle Sum Property :

$\\ \large \tt : \implies \: \angle dco + \angle odc + \angle doc = 180 \degree \\ \\ \large \tt : \implies \: \angle dco + 33 \degree + 110 \degree = 180 \degree \\ \\ \large \tt : \implies \: \angle dco = 180 \degree - 143 \degree \\ \\ \large \sf\purple{ \therefore \: \angle dco = 37 \degree}$

Now,

Angle DCO + Angle BCO = 90°

• Each angle of rectangle = 90°

$\\ \large \tt \implies \angle bco + 37 \degree = 90 \degree \\ \\ \large \tt \implies \: \angle bco = 90 \degree - 37 \degree \\ \\ \large \pink{ \therefore \: \angle bco = 53 \degree}$

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$\\ \large \mapsto \sf \: in \: \: \triangle abc :$

• Angle ACB = BCO = 53°
• Angle ACB = 90°
• Angle CAB = ?

⊙ Angle Sum Property :

$\\ \large \tt \implies \: \angle cab + 90 \degree + 53 \degree = 180 \degree \\ \\ \large \tt \implies \angle cab = 180 \degree - 143 \degree\\ \\ \large \underline{ \boxed{ \green{ \sf \therefore \: \angle oab = 37 \degree}}}$

Alternate method :

Angle OCD = Angle OAB

• (alternate interior angles)

Correct answer Angle OAB = 37°