Ten > Triangles and Quadrilaterals
In a circle with centre O, circumference angles RMP and RNP are drawn on the same arc RP. $\angle$ROP is the central angle.
- Write the relation between $\angle$RMP and $\angle$RNP.
- If $\angle$MRN=(7x-2)$^{\circ}$ and $\angle$MPN=(3x+10)$^{\circ}$, find the value of $\angle$MRN.
- Verify experimentally that the relation between $\angle$RMP and $\angle$ROP after drawing two circles having radii at least 3 cm.
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Atith Adhikari
·
1 year ago
Solution
In the given figure,
1) \( \rm \angle RMP = \angle RNP \). It is because they are the angles on the circumference of the same circle and standing on the same arc \( \rm {RP} \).
2) With the same reason as (1), \( \rm \angle MRN = \angle MPN \). Given,
\( \rm \angle MRN = (7x - 2)^{o} \ and \ \angle MPN = (3x + 10)^{o} \)
\( \rm (7x - 2) = (3x + 10) \)
\( \rm 7x - 2 + 2 = 3x + 10 + 2 \)
\( \rm 7x = 3x + 12 \)
\( \rm 7x - 3x = 3x - 3x + 12 \)
\( \rm 4x = 12 \)
\( \rm \therefore x = 3 \)
Hence, the required value of \( \rm \angle MRN = ( 7 \cdot 3 - 2 ) ^{o} = 19^{o} \).
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