Question

In the given figure, RS is the diameter of the circle. NM is parallel to RS and $\angle MRS={29}^o$. Calculate $\angle NRM$

• A. $\angle NRM={32}^o$
• B. $\angle NRM={42}^o$
• C. $\angle NRM={82}^o$
• D. $\angle NRM={12}^o$

Answer 1

The correct option is A

$\angle NRM={32}^o$

Join RN and MS

RS is the diameter

$\angle RMS={90}^o$ (angle in a semi circle)

$\angle RSM+\angle MRS={90}^0$

$\angle RSM={90}^o-{29}^o={61}^o$

But, $\angle RSM+\angle RNM={180}^0$ (Angles in a cyclic quadrilateral)

${61}^o+\angle RNM={180}^0$

$\angle RNM={180}^o-{61}^o={119}^o$

In Triangle NRM,

$\angle NRM+\angle RNM+\angle NMR={180}^o$

$\angle NRM+{119}^o+{29}^o={180}^o$

$\angle NRM+{148}^o={180}^o$

$\angle NRM={180}^o-{148}^o={32}^o$