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I think the events may be written as under 1st coin H T 2nd coin H. T 3rd coin H T Arrangement between 1st & 2nd coin H(H ) H(T) T(H ) T(T) In bracket the outcome of throwing 2nd coin Arrangement among 1st, 2nd & 3rd coin HH(H ) HT(H) TH(H) TT(H) HH(T ) HT(T) TH(T) TT(T) In bracket the outcome of throwing 3rd coin

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I like to think as below. There are 52 cards in a packet. Here are 2 events Event 1 To draw a red card other than kings . There are 26 red cards including 2 red kings Here are 24 red cards excluding two red kings Hence the probability to draw a red card other than king is 24/52=6/13 Event 2 To draw king There are 4 kings Hence the probability to draw a king is 4/52=1/13 Hence the Probability to draw a red card or a king is 6/13+1/13=7/13

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A solution without Trigonometry ---@@@---@@-@@@ Triangle PQR is a 30 60 90 triangle. Hence QR ঃ PQ =1ঃ√3 QR =PQ /√3 Triangle PQS is a 30 60 90 triangle Hence QSঃ PQ =√3ঃ1 > (QR+RS)ঃ PQ =√3ঃ1 >(PQ /√3+ 200)/PQ=√3 >1/√3+200/PQ=√3 >200/PQ=2/√3 >PQ =100√3=173.2 PQ=173.2 mitres Comment please.

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Inradius (r)=(P+B- H )/2

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AQ = AR=1/2(perimeter of 🔺 ABC )

PAB +PCB=PCB+x x =PAB=1/2*POB=1/2*100=50 degrees I think there is no need to find the magnitude of angle PCB

Join AC Triangle AOC AO=CO(radii) Angle ACO=angle BAC Angle BAC=angle x (circum angle on same arc) ang BAC + ang ACO=180-140=40 BAC=20 Hence angle x =20 degrees

Another method Exterior angle BOD =90+30=120 deg Triangle BOD is isosceles as OB =OD (those are radii) Angle BDO =angle =DBO=1/2(180-120)=30 deg

We may say that on arc BC the angle at centre (BOE) and angle on circumference (BDO) Hence angle BDO=1/2 angle BOE=60/2 degrees =30 degrees This will save time. Thanks. Comment please.

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This is one of the most under-rated video of secondary level mathematics.. This video deserves more.. This is a lengthy question based on the simple concept of rationalisation and during initial study ,students get confused about what to do and how to approach etc..

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