pls answer this question of both pls
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Answer:
Step-by-step explanation:
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this ans is of your new question
in\: \triangle PQR , PQ = PRin△PQR,PQ=PR
\angle {PQR} = \angle {PRQ} = 40\degree∠PQR=∠PRQ=40°
\blue {( Angles \: opposite \:to \: equal \:sides}(Anglesoppositetoequalsides
\blue{ equal. )}equal.)
\angle {QPR} + \angle {PQR} + \angle {PRQ}= 180\degree∠QPR+∠PQR+∠PRQ=180°
\orange { ( Angle \:Sum \: Property )}(AngleSumProperty)
\implies \angle {QPR} + 40\degree + 40 \:degree = 180\degree⟹∠QPR+40°+40degree=180°
\implies \angle {QPR} + 80\degree = 180\degree⟹∠QPR+80°=180°
\implies \angle {QPR} = 180\degree - 80\degree⟹∠QPR=180°−80°
\implies \angle {QPR} = 100\degree⟹∠QPR=100°
ii) Now, in\: \triangle QPS \:and \:in\: \triangle RPSii)Now,in△QPSandin△RPS
PQ = PR \: (given)PQ=PR(given)
\angle {PQR} = \angle {PRQ}∠PQR=∠PRQ
QS = SR \: ( \because PS \:is \:a \: median )QS=SR(∵PSisamedian)
\therefore \triangle QPS \cong \triangle RPS∴△QPS≅△RPS
\blue{ ( S .A.S \: congruence \:rule )}(S.A.Scongruencerule)
\angle {QPS} = \angle {RPS} \: ( C.P.CT )∠QPS=∠RPS(C.P.CT)
\implies \angle {QPS} = \frac{1}{2} \times \angle {QPR}⟹∠QPS=
2
1
×∠QPR
= \frac{1}{2} \times 100=
2
1
×100
= 50\degree=50°