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Given that mu = : 1 < x < 20, where x is an integer P = : x is a multiple of 3 \ and Q = : x is a prime number \ where P and Q are subsets of mu, find: (a) P Q'; (b) P' Q; (c) (P Q)'

Mathematics

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Answers

2 parts
a

129 m\text{129 m}

b

28.44 m\text{28.44 m}

Step-by-step solution

6 steps

Part a· 3 steps

  1. Identify the given values and the unknown height.

    We have a right-angled triangle TRO.
    The distance from the observer to the tower base, ∣OR∣|OR|, is 84 m.
    The angle of elevation from O to T is 57∘57^\circ.
    Let the height of the tower TR be hh.

  2. Use the tangent function to find the height hh.

    In △TRO\triangle TRO, tan⁡(θ)=oppositeadjacent\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}.

    tan⁡(57∘)=TROR\tan(57^\circ) = \frac{TR}{OR} tan⁡(57∘)=h84\tan(57^\circ) = \frac{h}{84} h=84×tan⁡(57∘)h = 84 \times \tan(57^\circ) h≈84×1.53986496h \approx 84 \times 1.53986496 h≈129.3486566h \approx 129.3486566
  3. Round the height to three significant figures.

    h≈129 mh \approx \boxed{\text{129 m}}

Part b

28.44 m\text{28.44 m}

  1. Identify the new angle and the unknown new distance.

    The observer moves away, and the new angle of elevation is 49∘49^\circ.
    Let the new distance from the tower base to the observer be OR′OR'.
    Using the height hh calculated in part (a) (using the unrounded value for accuracy):

    tan⁡(49∘)=hOR′\tan(49^\circ) = \frac{h}{OR'} OR′=htan⁡(49∘)OR' = \frac{h}{\tan(49^\circ)} OR′=129.3486566tan⁡(49∘)OR' = \frac{129.3486566}{\tan(49^\circ)} OR' \approx \frac{129.3486566}{1.15036736 OR′≈112.4400 mOR' \approx 112.4400 \text{ m}
  2. Calculate how far the observer moved backwards.

    The distance moved backwards is the difference between the new distance and the original distance.
    Distance moved =OR′−OR= OR' - OR
    Distance moved ≈112.4400−84\approx 112.4400 - 84
    Distance moved ≈28.4400 m\approx 28.4400 \text{ m}

  3. Round the distance moved to two decimal places.

    Distance moved ≈28.44 m\approx \boxed{\text{28.44 m}}

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