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Use the given sine ratio formula.

Mathematics

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9. In a right-angled triangle, the opposite side to an angle is 7 cm and the hypotenuse is 14 cm. Using the sine ratio (sin⁡θ=opposite/hypotenuse\sin \theta = \text{opposite/hypotenuse}), find the angle to the nearest degree.

  1. Use the given sine ratio formula.

    sin⁡θ=oppositehypotenuse\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}
  2. Substitute the given values into the formula.

    sin⁡θ=7 cm14 cm\sin \theta = \frac{7 \text{ cm}}{14 \text{ cm}}
  3. Simplify the fraction.

    sin⁡θ=12\sin \theta = \frac{1}{2}
  4. To find the angle θ\theta, take the inverse sine (arcsin) of 12\frac{1}{2}.

    θ=arcsin⁡(12)\theta = \arcsin\left(\frac{1}{2}\right) θ=30∘\theta = 30^\circ

    The angle to the nearest degree is 30∘\boxed{\text{}30^\circ\text{}}

    10. A surveyor needs to find the height of a building. She stands 50 metres away from the base of the building and measures the angle of elevation to the top as 35∘35^\circ.

  5. This problem forms a right-angled triangle where the height of the building is the opposite side to the angle of elevation, and the distance from the base is the adjacent side.

    We use the tangent ratio: tan⁡θ=oppositeadjacent\tan \theta = \frac{\text{opposite}}{\text{adjacent}}.
    Let hh be the height of the building.

    tan⁡35∘=h50 m\tan 35^\circ = \frac{h}{50 \text{ m}}
  6. Multiply both sides by 50 m to solve for hh.

    h=50 m×tan⁡35∘h = 50 \text{ m} \times \tan 35^\circ
  7. Calculate the value of tan⁡35∘\tan 35^\circ and then multiply.

    h≈50 m×0.7002h \approx 50 \text{ m} \times 0.7002 h≈35.01 mh \approx 35.01 \text{ m}

    The height of the building is approximately 35.0 m\boxed{\text{}35.0 \text{ m}\text{}}

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