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Two customer service agents in a bank work such that the first one takes an average of 5 minutes while the second takes an average of 8 minutes to serve a client. The counters are opened at 8:00 am. When the agents begin serving clients -At what time will they complete serving 182 clients?

Mathematics

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Answer

2424.39 m2424.39 \text{ m}

Step-by-step solution

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You're on a roll —Ili kutatua swali hili, tutatumia kanuni za trigonometry.

  1. Chora mchoro wa hali ilivyo.

    Hebu HH iwe urefu wa mlima.
    Hebu xx iwe umbali wa usawa kutoka sehemu P hadi chini ya mlima.
    Hebu VV iwe sehemu ya kwanza ya uchunguzi.
    Hebu PP iwe sehemu ya pili ya uchunguzi.
    Umbali kutoka V hadi chini ya mlima ni x+2000x + 2000 m.
    Pembe ya kuinuka kutoka V ni 22∘22^\circ.
    Pembe ya kuinuka kutoka P ni 40∘40^\circ.
    Tunatafuta umbali mfupi zaidi kutoka P hadi kilele cha mlima, ambao ni hypotenuse ya pembetatu ya kulia iliyoundwa na P, chini ya mlima, na kilele cha mlima. Hebu tuite umbali huu LL.

  2. Andika milinganyo ya trigonometric.

    Kutoka sehemu V:

    tan⁡(22∘)=Hx+2000\tan(22^\circ) = \frac{H}{x + 2000} H=(x+2000)tan⁡(22∘)(Mlinganyo1)H = (x + 2000)\tan(22^\circ) \quad (Mlinganyo 1)

    Kutoka sehemu P:

    tan⁡(40∘)=Hx\tan(40^\circ) = \frac{H}{x} H=xtan⁡(40∘)(Mlinganyo2)H = x\tan(40^\circ) \quad (Mlinganyo 2)
  3. Tatua kwa xx (umbali wa usawa kutoka P).

    Linganisha Mlinganyo 1 na Mlinganyo 2:

    (x+2000)tan⁡(22∘)=xtan⁡(40∘)(x + 2000)\tan(22^\circ) = x\tan(40^\circ)

    Panua upande wa kushoto:

    xtan⁡(22∘)+2000tan⁡(22∘)=xtan⁡(40∘)x\tan(22^\circ) + 2000\tan(22^\circ) = x\tan(40^\circ)

    Panga upya ili kutenga xx:

    2000tan⁡(22∘)=xtan⁡(40∘)−xtan⁡(22∘)2000\tan(22^\circ) = x\tan(40^\circ) - x\tan(22^\circ) 2000tan⁡(22∘)=x(tan⁡(40∘)−tan⁡(22∘))2000\tan(22^\circ) = x(\tan(40^\circ) - \tan(22^\circ))

    Tatua kwa xx:

    x=2000tan⁡(22∘)tan⁡(40∘)−tan⁡(22∘)x = \frac{2000\tan(22^\circ)}{\tan(40^\circ) - \tan(22^\circ)}

    Tumia thamani za tan: tan⁡(22∘)≈0.4040\tan(22^\circ) \approx 0.4040 na tan⁡(40∘)≈0.8391\tan(40^\circ) \approx 0.8391.

    x=2000×0.40400.8391−0.4040x = \frac{2000 \times 0.4040}{0.8391 - 0.4040} x=8080.4351x = \frac{808}{0.4351} x≈1857.04 mx \approx 1857.04 \text{ m}
  4. Tatua kwa HH (urefu wa mlima).

    Tumia Mlinganyo 2 na thamani ya xx:

    H=xtan⁡(40∘)H = x\tan(40^\circ) H=1857.04×0.8391H = 1857.04 \times 0.8391 H≈1558.30 mH \approx 1558.30 \text{ m}
  5. Kokotoa umbali mfupi zaidi kutoka P hadi kilele cha mlima (LL).

    Umbali mfupi zaidi ni hypotenuse ya pembetatu ya kulia iliyoundwa na HH na xx. Tunaweza kutumia sin⁡(40∘)=HL\sin(40^\circ) = \frac{H}{L}.

    L=Hsin⁡(40∘)L = \frac{H}{\sin(40^\circ)}

    Tumia thamani ya sin⁡(40∘)≈0.6428\sin(40^\circ) \approx 0.6428.

    L=1558.300.6428L = \frac{1558.30}{0.6428} L≈2424.39 mL \approx 2424.39 \text{ m}

    Umbali mfupi zaidi kutoka sehemu P hadi kilele cha mlima ni:

    2424.39 m2424.39 \text{ m}
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