Answer:
m∠AOD = 140°
Explanation:
In the diagram attached,
OA⊥OC and OB⊥OD
m∠AOD = 3.5(m∠BOC)
Since, m∠BOD = 90° [Given: OA⊥OC]
m∠BOC + m∠COD = 90° ---------(1)
Similarly, m∠AOC = 90° [Given : OA⊥OC]
m∠AOB + m∠BOC = 90° --------(2)
Equation (1) - Equation(2)
(m∠BOC + m∠COD) - (m∠AOB + m∠BOC) = 90°- 90°
m∠COD = m∠AOB
m∠AOB + m∠BOC + m∠COD = m∠AOD --------(3)
m∠AOB + m∠BOC + m∠AOB = 3.5(m∠BOC) [Since m∠COD = m∠AOB]
2m∠AOB = 3.5(m∠BOC) - m∠BOC
2m∠AOB = 2.5(m∠BOC)
m∠AOB = 1.25(m∠BOC)
From equation (2),
m∠AOB + m∠BOC = 90°
1.25(m∠BOC) + m∠BOC = 90°
2.25(m∠BOC) = 90°
m∠BOC = 40°
From equation (1),
m∠BOC + m∠COD = 90°
m∠COD + 40° = 90°
m∠COD = 50°
Now by putting these values in equation (3)
m∠AOB + m∠BOC + m∠COD = m∠AOD
m∠COD + m∠BOC + m∠COD = m∠AOD
50° + 40° + 50°= m∠AOD
m∠AOD = 140°