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What are the answers to these trigonometry problems involving angles of elevation, bearings, and finding the length of a side in a right-angled triangle?

Mathematics

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here are the solutions to the problems:

1. Angle of elevation and depression are measured from the (horizontal/vertical)

Angles of elevation and depression are always measured from the horizontal line.
Answer:
Angles of elevation and depression are measured from the horizontal\boxed{\mathbf{horizontal}}.

2. Write down the bearing of P from A.

  1. Understand the definition of bearing.

    Bearings are measured clockwise from the North line, usually expressed as a three-digit number.

  2. Read the angle from the diagram.

    The diagram shows a North line at point A. The angle from the North line clockwise to the line segment AP is given as 56∘56^\circ.

  3. Express the bearing in three digits.

    Since 56∘56^\circ is a two-digit number, we add a leading zero to make it three digits.
    Answer:
    The bearing of P from A is 056∘\boxed{\mathbf{056^\circ}}.

    3. Find the exact length of AC.

  4. Identify the given information in the right-angled triangle ABC.

    Angle C is 90∘90^\circ.
    Angle B is 30∘30^\circ.
    Length of BC (adjacent side to angle B) is 12.
    We need to find the length of AC (opposite side to angle B).

  5. Choose the appropriate trigonometric ratio.

    Since we have the opposite side and the adjacent side relative to angle B, we use the tangent function:

    tan⁡(θ)=oppositeadjacent\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
  6. Substitute the known values into the formula.

    tan⁡(30∘)=AC12\tan(30^\circ) = \frac{AC}{12}
  7. Solve for AC.

    AC=12×tan⁡(30∘)AC = 12 \times \tan(30^\circ)
  8. Use the exact value of tan⁡(30∘)\tan(30^\circ).

    The exact value of tan⁡(30∘)\tan(30^\circ) is 13\frac{1}{\sqrt{3}} or 33\frac{\sqrt{3}}{3}.

    AC=12×33AC = 12 \times \frac{\sqrt{3}}{3} AC=43AC = 4\sqrt{3}

    Answer:
    The exact length of AC is 43\boxed{\mathbf{4\sqrt{3}}}.

    4. Q is due East of R and due South of P. The bearing of R from P is 230°. Show this information on the diagram.

    To show this information on a diagram, you would draw the points and lines as follows:

  9. Place point P.

    Draw a North line from P.

  10. From the North line at P, measure 230∘230^\circ clockwise.

    Draw a line segment in this direction. Point R will be located somewhere along this line segment.

  11. Since Q is due South of P, draw a vertical line downwards from P.

    Point Q will be located on this line.

  12. Since Q is due East of R, draw a horizontal line to the right from R.

    Point Q will be located on this line.

  13. The intersection of the vertical line from P and the horizontal line from R will be point Q.

    Ensure that Q is to the right of R and below P.

    Diagram Description:

    • Draw point P.
    • Draw a vertical line upwards from P, labeled 'N' for North.
    • From the North line at P, draw a line segment PR such that the angle measured clockwise from the North line to PR is 230∘230^\circ.
    • Draw a vertical line downwards from P. Point Q lies on this line.
    • Draw a horizontal line to the right from R. Point Q lies on this line.
    • The point where the vertical line from P and the horizontal line from R intersect is point Q.
    • This arrangement will show P, Q, and R forming a right-angled triangle with the right angle at Q, where PQ is a vertical line and RQ is a horizontal line.

    Answer:
    The information is shown by drawing points P, Q, and R such that:

    • A line from P to R makes an angle of 230∘230^\circ clockwise from the North line at P.
    • Q is directly below P (on a vertical line).
    • Q is directly to the right of R (on a horizontal line).
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