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A ladder 10m long is leaning against a wall. The bottom of the ladder is 3m away from the wall. The wall is 26m high. Draw a sketch and calculate the length of the ladder touching the wall.

Mathematics

ScanSolve AI Answer

Step-by-step solution

11 steps

Part i:

a) Draw a sketch of the plot.

  1. Draw a right-angled triangle ABC with the right angle at A.

    Mark side AB as 10 meters and side AC as 26 meters.
    Locate point D on side AB such that BD = 3 meters. This means AD = AB - BD = 10 - 3 = 7 meters.
    Draw a line segment from C to D, and label the angle ACD as θ\theta.

    \begin{tikzpicture}[scale=0.6] \coordinate (A) at (0,0); \coordinate (B) at (10,0); \coordinate (C) at (0,26); \coordinate (D) at (7,0); % D is on AB, AD = 7, DB = 3 \draw (A) -- (B) node[midway, below] {10 m} -- (C) -- cycle; \draw (A) -- (C) node[midway, left] {26 m}; \draw (C) -- (D); \node at (A) [below left] {A}; \node at (B) [below right] {B}; \node at (C) [above left] {C}; \node at (D) [below] {D}; \draw (0.5,0) -- (0.5,0.5) -- (0,0.5); % Right angle at A \node at (D) [below] {D}; \node at (B) [below] {B}; \node at (A) [below] {A}; \draw (D) -- (A) node[midway, below] {7 m}; \draw (B) -- (D) node[midway, below] {3 m}; \draw pic["$\theta$", draw, angle radius=1cm] {angle = D--C--A}; \end{tikzpicture}

    b) Help him to get the length of the side BC.

  2. Use the Pythagorean theorem for the right-angled triangle ABC.

    BC2=AB2+AC2BC^2 = AB^2 + AC^2

  3. Substitute the given values for AB and AC.

    BC2=(10 m)2+(26 m)2BC^2 = (10\, \text{m})^2 + (26\, \text{m})^2 BC2=100 m2+676 m2BC^2 = 100\, \text{m}^2 + 676\, \text{m}^2 BC2=776 m2BC^2 = 776\, \text{m}^2

  4. Calculate the square root to find BC.

    BC=776 mBC = \sqrt{776}\, \text{m} BC≈27.8567 mBC \approx 27.8567\, \text{m} BC≈27.86 mBC \approx \boxed{27.86\, \text{m}}

    c) Calculate the value of θ\theta to two decimal places.

  5. Identify the relevant right-angled triangle, which is ADC.

    In △ADC\triangle ADC, the right angle is at A.
    AD = 7 m (since D is 3 m from B along AB, and AB = 10 m, so AD = 10 - 3 = 7 m).
    AC = 26 m.

  6. Use the tangent function to find θ\theta.

    tan⁡(θ)=OppositeAdjacent=ADAC\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AD}{AC}

  7. Substitute the values and calculate θ\theta.

    tan⁡(θ)=726\tan(\theta) = \frac{7}{26} θ=arctan⁡(726)\theta = \arctan\left(\frac{7}{26}\right) θ≈15.0699∘\theta \approx 15.0699^\circ θ≈15.07∘\theta \approx \boxed{15.07^\circ}

    Part ii:

    a) Using the bar chart, copy and complete the following table.

  8. Read the number of bottles for each color from the bar chart.

    • Red: 25 bottles
    • Blue: 15 bottles
    • Green: 28 bottles
    • Yellow: 20 bottles
    • Black: 8 bottles
    • White: 11 bottles
    ColorRedBlueGreenYellowBlackWhite
    Number of bottles25152820811

    b) Find the total number of bottles of coloring materials in the workshop.

  9. Sum the number of bottles for all colors.

    Total bottles=25+15+28+20+8+11\text{Total bottles} = 25 + 15 + 28 + 20 + 8 + 11

  10. Perform the addition.

    Total bottles=107\text{Total bottles} = 107

    The total number of bottles is 107\boxed{107}.

    c) What is the modal color?

  11. Identify the color with the highest number of bottles from the table in part a).

    The highest number of bottles is 28, which corresponds to Green.
    The modal color is Green\boxed{\text{Green}}.

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