Question

ABCD is a cyclic quadrilateral. Given that $\angle ADB+\angle DAB={120}^o$ and $\angle ABC+\angle BDA={145}^o$. Find the value of $\angle CDB$.

• A. ${75}^o$
• B. ${115}^o$
• C. ${35}^o$
• D. ${45}^o$

Answer 1

The correct option is C ${35}^o$

We have;

$\angle ADB+\angle DAB=120°$ (1)

$\angle ABC+\angle BDA=145°$ (2)

$\angle BDA+\angle CDB+\angle ABC=180°$ (3) [ In cyclic quadrilateral, sum of diagonal angles is 180°]

Subtracting equation 2 from equation 3, we get;

$\angle CDB=35°$

Hence, $\angle CDB=35°$