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Identify the known values in the right-angled triangle BFC.

Mathematics

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2 parts
a

25.35 m25.35 \text{ m}

b

60.40∘60.40^\circ

187.5 kg/m3187.5 \text{ kg/m}^3

41.08 cm41.08 \text{ cm}

Step-by-step solution

17 steps

50.

Part a· 4 steps

The height of the tree

  1. Identify the known values in the right-angled triangle BFC.

    The distance from the foot of the tree (F) to the chameleon (C) is FC=14.4 mFC = 14.4 \text{ m}.
    The angle of elevation of the bird (B) from the chameleon (C) is ∠BCF=41.4∘\angle BCF = 41.4^\circ.

  2. Use the tangent ratio to find the height of the bird (BF).

    tan⁡(∠BCF)=BFFC\tan(\angle BCF) = \frac{BF}{FC} tan⁡(41.4∘)=BF14.4 m\tan(41.4^\circ) = \frac{BF}{14.4 \text{ m}} BF=14.4 m×tan⁡(41.4∘)BF = 14.4 \text{ m} \times \tan(41.4^\circ) BF=14.4 m×0.88009BF = 14.4 \text{ m} \times 0.88009 BF≈12.6733 mBF \approx 12.6733 \text{ m}

  3. Calculate the height of the tree (TF).

    Since B is the midpoint of FT, the height of the tree TF is twice the height of the bird BF.

    TF=2×BFTF = 2 \times BF TF=2×12.6733 mTF = 2 \times 12.6733 \text{ m} TF≈25.3466 mTF \approx 25.3466 \text{ m}

  4. Round the height to 2 decimal places.

    TF≈25.35 mTF \approx 25.35 \text{ m}

    The height of the tree is 25.35 m\boxed{25.35 \text{ m}}.

Part b

60.40∘60.40^\circ

The angle of depression of the chameleon from point T

  1. Identify the known values in the right-angled triangle TFC.

    The height of the tree TF≈25.3466 mTF \approx 25.3466 \text{ m} (from part a).
    The distance from the foot of the tree (F) to the chameleon (C) is FC=14.4 mFC = 14.4 \text{ m}.

  2. The angle of depression of the chameleon from point T is equal to the angle of elevation of T from C, which is ∠TCF\angle TCF.

    Use the tangent ratio.

    tan⁡(∠TCF)=TFFC\tan(\angle TCF) = \frac{TF}{FC} tan⁡(∠TCF)=25.3466 m14.4 m\tan(\angle TCF) = \frac{25.3466 \text{ m}}{14.4 \text{ m}} tan⁡(∠TCF)≈1.76018\tan(\angle TCF) \approx 1.76018

  3. Calculate the angle ∠TCF\angle TCF.

    ∠TCF=arctan⁡(1.76018)\angle TCF = \arctan(1.76018) ∠TCF≈60.399∘\angle TCF \approx 60.399^\circ

  4. Round the angle to 2 decimal places.

    ∠TCF≈60.40∘\angle TCF \approx 60.40^\circ

    The angle of depression of the chameleon from point T is 60.40∘\boxed{60.40^\circ}.

    51.

  5. Convert the side length of the cubical stone from cm to m.

    s=40 cm=0.4 ms = 40 \text{ cm} = 0.4 \text{ m}

  6. Calculate the volume of the cubical stone in m3^3.

    V=s3V = s^3 V=(0.4 m)3V = (0.4 \text{ m})^3 V=0.064 m3V = 0.064 \text{ m}^3

  7. Use the given mass and calculated volume to determine the density.

    m=12 kgm = 12 \text{ kg} ρ=mV\rho = \frac{m}{V} ρ=12 kg0.064 m3\rho = \frac{12 \text{ kg}}{0.064 \text{ m}^3} ρ=187.5 kg/m3\rho = 187.5 \text{ kg/m}^3

    The density of the stone is 187.5 kg/m3\boxed{187.5 \text{ kg/m}^3}.

    52.

  8. Use the area formula for a triangle to find ∠PQR\angle PQR.

    Given Area =303 cm2= 30\sqrt{3} \text{ cm}^2, PQ=10 cmPQ = 10 \text{ cm}, QR=12 cmQR = 12 \text{ cm}.

    Area=12×PQ×QR×sin⁡(∠PQR)\text{Area} = \frac{1}{2} \times PQ \times QR \times \sin(\angle PQR) 303=12×10×12×sin⁡(∠PQR)30\sqrt{3} = \frac{1}{2} \times 10 \times 12 \times \sin(\angle PQR) 303=60×sin⁡(∠PQR)30\sqrt{3} = 60 \times \sin(\angle PQR) sin⁡(∠PQR)=30360\sin(\angle PQR) = \frac{30\sqrt{3}}{60} sin⁡(∠PQR)=32\sin(\angle PQR) = \frac{\sqrt{3}}{2}

  9. Determine the obtuse angle ∠PQR\angle PQR.

    Since sin⁡(∠PQR)=32\sin(\angle PQR) = \frac{\sqrt{3}}{2}, the acute angle is 60∘60^\circ. As ∠PQR\angle PQR is obtuse,

    ∠PQR=180∘−60∘=120∘\angle PQR = 180^\circ - 60^\circ = 120^\circ

  10. Use the Cosine Rule to find the length of PR.

    PR2=PQ2+QR2−2×PQ×QR×cos⁡(∠PQR)PR^2 = PQ^2 + QR^2 - 2 \times PQ \times QR \times \cos(\angle PQR) PR2=(10 cm)2+(12 cm)2−2×10 cm×12 cm×cos⁡(120∘)PR^2 = (10 \text{ cm})^2 + (12 \text{ cm})^2 - 2 \times 10 \text{ cm} \times 12 \text{ cm} \times \cos(120^\circ) PR2=100 cm2+144 cm2−240 cm2×(−12)PR^2 = 100 \text{ cm}^2 + 144 \text{ cm}^2 - 240 \text{ cm}^2 \times \left(-\frac{1}{2}\right) PR2=244 cm2+120 cm2PR^2 = 244 \text{ cm}^2 + 120 \text{ cm}^2 PR2=364 cm2PR^2 = 364 \text{ cm}^2 PR=364 cmPR = \sqrt{364} \text{ cm} PR≈19.0787 cmPR \approx 19.0787 \text{ cm}

  11. Calculate the perimeter of the triangle.

    Perimeter=PQ+QR+PR\text{Perimeter} = PQ + QR + PR Perimeter=10 cm+12 cm+19.0787 cm\text{Perimeter} = 10 \text{ cm} + 12 \text{ cm} + 19.0787 \text{ cm} Perimeter=41.0787 cm\text{Perimeter} = 41.0787 \text{ cm}

  12. Round the perimeter to 2 decimal places.

    Perimeter≈41.08 cm\text{Perimeter} \approx 41.08 \text{ cm}

    The perimeter of the triangle is 41.08 cm\boxed{41.08 \text{ cm}}.

    53.

  13. Calculate the total time elapsed from Monday 6:00 a.m.

    to Friday 6:00 a.m.
    From Monday 6:00 a.m. to Friday 6:00 a.m. is 4 full days.
    Total hours = 4 days×24 hours/day=96 hours4 \text{ days} \times 24 \text{ hours/day} = 96 \text{ hours}.
    Step

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