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Solve the following trigonometry problems: 1. A 13m ladder leans against a wall, its base is 5m from the wall. How high up the wall does it reach? 2. From a cliff 15m high, the angle of depression to

Mathematics

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Step-by-step solution

14 steps

Here are the solutions to the trigonometry problems:

: Ladder against a wall

  1. Identify the knowns and unknowns.

    The ladder, the wall, and the ground form a right-angled triangle.

    • Length of the ladder (hypotenuse) = 13 m
    • Distance from the base of the wall to the foot of the ladder (adjacent side) = 5 m
    • Height the ladder reaches up the wall (opposite side) = hh (unknown)
  2. Apply the Pythagorean theorem.

    The Pythagorean theorem states a2+b2=c2a^2 + b^2 = c^2, where cc is the hypotenuse and aa and bb are the other two sides.

    h2+52=132h^2 + 5^2 = 13^2
  3. Solve for hh.

    h2+25=169h^2 + 25 = 169 h2=169−25h^2 = 169 - 25 h2=144h^2 = 144 h=144h = \sqrt{144} h=12h = 12

    The ladder reaches 12 m\boxed{\text{12 m}} up the wall.

    : Angle of depression of a boat

  4. Identify the knowns and unknowns.

    The cliff, the sea level, and the line of sight to the boat form a right-angled triangle. The angle of depression from the top of the cliff to the boat is equal to the angle of elevation from the boat to the top of the cliff.

    • Height of the cliff (opposite side to the angle of elevation) = 15 m
    • Angle of depression = 35∘35^\circ (which is also the angle of elevation from the boat)
    • Distance from the base of the cliff to the boat (adjacent side) = xx (unknown)
  5. Use the tangent function.

    The tangent of an angle in a right-angled triangle is the ratio of the opposite side to the adjacent side:

    tan⁡(θ)=oppositeadjacent\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} tan⁡(35∘)=15x\tan(35^\circ) = \frac{15}{x}
  6. Solve for xx.

    We are given tan⁡35∘=0.7\tan 35^\circ = 0.7.

    0.7=15x0.7 = \frac{15}{x} x=150.7x = \frac{15}{0.7} x≈21.42857x \approx 21.42857
  7. Round to two decimal places.

    x≈21.43x \approx 21.43

    The distance from the base of the cliff to the boat is 21.43 m\boxed{\text{21.43 m}}.

    : Angle of depression to a swimmer

  8. Identify the knowns and unknowns.

    The cliff, the sea level, and the line of sight to the swimmer form a right-angled triangle.

    • Height of the cliff (opposite side) = 30 m
    • Horizontal distance to the swimmer (adjacent side) = 15 m
    • Angle of depression = θ\theta (unknown)
  9. Use the tangent function.

    tan⁡(θ)=oppositeadjacent\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} tan⁡(θ)=3015\tan(\theta) = \frac{30}{15} tan⁡(θ)=2\tan(\theta) = 2
  10. Determine the angle using the given values.

    We are given:
    tan⁡64∘=2.0503\tan 64^\circ = 2.0503
    tan⁡65∘=2.1444\tan 65^\circ = 2.1444
    tan⁡66∘=2.2460\tan 66^\circ = 2.2460
    Since tan⁡(θ)=2\tan(\theta) = 2, the angle θ\theta is between 63∘63^\circ and 64∘64^\circ. Based on the provided options, the closest integer angle would be 63∘63^\circ if we were to round, but since 2 is not exactly one of the given values, we can state the exact value of θ\theta.

    θ=arctan⁡(2)\theta = \arctan(2)

    Using a calculator, θ≈63.43∘\theta \approx 63.43^\circ.
    However, if we are to choose from the given values, none of them exactly match tan⁡(θ)=2\tan(\theta) = 2. If the question implies finding the closest integer angle from the given table, then 2 is closest to 2.0503 (for 64∘64^\circ) than to 2.1444 (for 65∘65^\circ). But 2 is also less than 2.0503.
    Let's assume the question expects us to find the angle whose tangent is exactly 2.
    The angle of depression is arctan⁡(2)≈63.43∘\boxed{\arctan(2) \approx 63.43^\circ}.

    : Height of a tree

  11. Identify the knowns and unknowns.

    We have a right-angled triangle formed by Bill's eye level, the horizontal distance to the tree, and the height from Bill's eye level to the top of the tree.

    • Bill's height = 1.5 m
    • Horizontal distance from Bill to the tree (adjacent side) = 5 m
    • Angle of elevation = 20∘20^\circ
    • Height from Bill's eye level to the top of the tree = h′h' (unknown)
  12. Use the tangent function to find h′h'.

    tan⁡(θ)=oppositeadjacent\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} tan⁡(20∘)=h′5\tan(20^\circ) = \frac{h'}{5}
  13. Solve for h′h'.

    We are given tan⁡20∘=0.36\tan 20^\circ = 0.36.

    0.36=h′50.36 = \frac{h'}{5} h′=0.36×5h' = 0.36 \times 5

    h′=1.8h' = 1.8 m

  14. Calculate the total height of the tree.

    The total height of the tree is Bill's height plus h′h'.
    Total height = Bill's height + h′h'
    Total height = 1.5 m+1.8 m1.5 \text{ m} + 1.8 \text{ m}
    Total height = 3.3 m
    The height of the tree is 3.3 m\boxed{\text{3.3 m}}.

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