Help~STPM Trigonometry & Probability

kitwei1219

1)If A+B+C=90,prove that cotA + cotB + cotC = cotAcotBcotC

2)The events A,B,C and D are mutually exclusive and P(A)=P(B)=P(C)=P(D)=1/6.If the event E=A U B U C and the event F=C U D,find P(E ∩ F) and P(E U F).


3)At a dinner,15 bottled drinks were opened.There were 7 bottles of orange juice,3 of which are brown bottles,and the remaining were green bottles. There were 8 bottles of lemonade,2 of which are brown while the rest were green.These bottles were opened,one by one,in random order.The events A,Band C are defined as follows.
A:The first 2 bottles opened are brown.
B:The first 2 bottles opened contain lemonade.
C:The last 2 bottles opened are green and contain orange juice.
Find the values of P(A),P(C),P(BIA),P(CIA) and P(A ∩ B'),where B' is the event 'not B'.


4)In Brobdignag,the weather each day is either fine or rainy.If the weather is fine,then the probability that the following day will also have fine weather is 0.7.If the weather is rainy,then the probability that the following day will also be rainy is 0.8.
(i)Suppose that it is known that the weather on 1 August will be fine.Show that the probabolity that the weather on 3 August will be fine is 0.55,and the probability that the weather on 5 August will be fine.
(ii)Given that there is a probabilty if 0.4 that the weather on 1 September is fine,find the probabilty that the there will be fine weather on2 September,and find also the probabiity that there will be fine weather on 3 September.


ans:2)1/6;2/3
       3)2/21,2/35,1/10,1/13,3/35
       4)(i)0.4375  (ii)0.4,0.4


恳求各位的帮忙。谢谢。

Mouse_Girl

2)The events A,B,C and D are mutually exclusive and P(A)=P(B)=P(C)=P(D)=1/6.If the event E=A U B U C and the event F=C U D,find P(E ∩ F) and P(E U F)

E ∩ F = (A U B U C) ∩ (C U D) = C
So, P(E ∩ F) = P(C) = 1/6

E U F = A U B U C U D
So, P(E U F) = 1/6 + 1/6 + 1/6 + 1/6 = 2/3

kitwei1219

哈哈，谢谢。XD

~HeBe~_@

原帖由 kitwei1219 于 7-10-2009 07:19 PM 发表
1)If A+B+C=90,prove that cotA + cotB + cotC = cotAcotBcotC

Solution:

As we know
    A+B+C=90°
=>    A+B = 90° - C
=> sin (A+B) = sin (90° - C) = cos C
or
=> cos(A+B) = cos (90° - C) = sin C

then,
LHS = cotA               + cotB               + cotC
        = (cosA)/(sinA) + (cosB)/(sinB) + (cosC)/(sinC)
        = (cosA.sinB.sinC + cosB.sinA.sinC + cosC.sinA.sinB)/(sinAsinBsinC)
        = [sinC(cosA.sinB + cosB.sinA) + cosC.sinA.sinB]/(sinAsinBsinC)
        = [sinC(sinA.cosB + cosA.sinB) + cosC.sinA.sinB]/(sinAsinBsinC)
        = [sinC. sin(A+B)                         +cosC.sinA.sinB]/(sinAsinBsinC)
        = [sinC. cosC                               +cosC.sinA.sinB]/(sinAsinBsinC)
        = cosC [sinC         +sinA.sinB]/(sinAsinBsinC)
        = cosC [cos(A+B) +sinA.sinB]/(sinAsinBsinC)
        = cosC [cosA.cosB - sinA.sinB +sinA.sinB]/(sinAsinBsinC)
        = cosC [cosA.cosB]/(sinAsinBsinC)
        = (cosA.cosB.cosC)/(sinAsinBsinC)
        = cotA.cotB.cotC     (proven)

kitwei1219

谢谢

mathlim

1) If A+B+C=90°, prove that cotA + cotB + cotC = cotAcotBcotC

tan(A+B) = tan(90°-C) = cotC = 1/tanC

(tanA+tanB)/(1-tanAtanB) = 1/tanC

tanAtanC+tanBtanC = 1-tanAtanB

tanBtanC+tanAtanC+tanAtanB = 1

(tanBtanC+tanAtanC+tanAtanB)/(tanAtanBtanC) = 1/(tanAtanBtanC)

cotA + cotB + cotC = cotAcotBcotC

kitwei1219

mathlim好强啊～佩服佩服。再次谢谢你。