In the given figure (not to scale), $P Q | | R S, ∠ Q P M = 30^{o}, ∠ S R T = 85^{o}$ and $R M = R P$ , then $∠ R M N$ is equal to

90^{o}

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110^{o}

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115^{o}

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125^{o}

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Solution

The correct option is B $110^{o}$
$P Q | | R S$
$∴ Q P R =< ∠ S R T = 85^{0}$
$∴ ∠ M P R = 85 - 30 = 55^{0}$
$∠ M R P = ∠ M P R = 55^{0}$ (isosceles $Δ l e$ )
$∠ R M N = ∠ M P R + ∠ M R P$ (external angle thorem)
$= 55 + 55 = 110^{0}$

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In the given figure (not to scale), PQ||RS,∠QPM=30o,∠SRT=85o and RM=RP, then ∠RMN is equal to

The correct option is B 110oPQ||RS∴QPR=<∠SRT=850∴∠MPR=85−30=550∠MRP=∠MPR=550 (isosceles Δle )∠RMN=∠MPR+∠MRP (external angle thorem)=55+55=1100

In the given figure (not to scale), $P Q | | R S, ∠ Q P M = 30^{o}, ∠ S R T = 85^{o}$ and $R M = R P$ , then $∠ R M N$ is equal to

The correct option is B $110^{o}$
$P Q | | R S$
$∴ Q P R =< ∠ S R T = 85^{0}$
$∴ ∠ M P R = 85 - 30 = 55^{0}$
$∠ M R P = ∠ M P R = 55^{0}$ (isosceles $Δ l e$ )
$∠ R M N = ∠ M P R + ∠ M R P$ (external angle thorem)
$= 55 + 55 = 110^{0}$