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Use the Sine Rule.

Mathematics

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37. In a triangle, if a=10 cma = 10 \text{ cm}, A=30∘A = 30^\circ, and B=60∘B = 60^\circ, find bb.

  1. Use the Sine Rule.

    The Sine Rule states that for a triangle with sides a,b,ca, b, c and opposite angles A,B,CA, B, C:

    asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
  2. Substitute the given values into the Sine Rule.

    10 cmsin⁡30∘=bsin⁡60∘\frac{10 \text{ cm}}{\sin 30^\circ} = \frac{b}{\sin 60^\circ}
  3. Solve for bb.

    We know that sin⁡30∘=0.5\sin 30^\circ = 0.5 and sin⁡60∘=32≈0.866\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866.

    100.5=b0.866\frac{10}{0.5} = \frac{b}{0.866} 20=b0.86620 = \frac{b}{0.866} b=20×0.866b = 20 \times 0.866 b=17.32 cmb = 17.32 \text{ cm}
  4. Compare with the given options.

    The calculated value 17.32 cm17.32 \text{ cm} is closest to option B.

    The final answer is B. 17.3 cm\boxed{\text{B. } 17.3 \text{ cm}}.

    38. Which formula is the cosine rule?

    The Cosine Rule relates the lengths of the sides of a triangle to the cosine of one of its angles. For a triangle with sides a,b,ca, b, c and angle AA opposite side aa, the formula is:

    a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc \cos A

    Comparing this with the given options:
    A. a2=b2+c2a^2 = b^2 + c^2 is the Pythagorean theorem.
    B. a2=b2+c2−2bccos⁡Aa^2 = b^2 + c^2 - 2bc \cos A is the Cosine Rule.
    C. a/sin⁡A=b/sin⁡Ba/\sin A = b/\sin B is the Sine Rule.
    D. tan⁡A=opposite/adjacent\tan A = \text{opposite}/\text{adjacent} is a trigonometric ratio for right-angled triangles.

    The final answer is B. a2=b2+c2−2bccos⁡A\boxed{\text{B. } a^2 = b^2 + c^2 - 2bc \cos A}.

    39. A ladder 5 m5 \text{ m} long leans against a wall, forming a 60∘60^\circ angle with the ground. What is the height the ladder reaches on the wall?

  5. Identify the components of the right-angled triangle formed.

    The ladder is the hypotenuse (H=5 mH = 5 \text{ m}).
    The angle with the ground is θ=60∘\theta = 60^\circ.
    The height the ladder reaches on the wall is the side opposite to the angle (Opposite=h\text{Opposite} = h).

  6. Use the sine function.

    The sine function relates the opposite side, hypotenuse, and angle:

    sin⁡θ=OppositeHypotenuse\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}
  7. Substitute the values and solve for hh.

    sin⁡60∘=h5 m\sin 60^\circ = \frac{h}{5 \text{ m}} h=5 m×sin⁡60∘h = 5 \text{ m} \times \sin 60^\circ

    We know that sin⁡60∘=32≈0.866\sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866.

    h=5 m×0.866h = 5 \text{ m} \times 0.866 h=4.33 mh = 4.33 \text{ m}
  8. Compare with the given options.

    The calculated value 4.33 m4.33 \text{ m} is closest to option C.

    The final answer is C. 4.3 m\boxed{\text{C. } 4.3 \text{ m}}.

    40. A man sees the top of a building, an angle of elevation of 30∘30^\circ. He is standing 20 m20 \text{ m} from the foot of the building. What is the height of the building?

  9. Identify the components of the right-angled triangle formed.

    The distance from the building is the adjacent side (Adjacent=20 m\text{Adjacent} = 20 \text{ m}).
    The angle of elevation is θ=30∘\theta = 30^\circ.
    The height of the building is the side opposite to the angle (Opposite=h\text{Opposite} = h).

  10. Use the tangent function.

    The tangent function relates the opposite side, adjacent side, and angle:

    tan⁡θ=OppositeAdjacent\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}}
  11. Substitute the values and solve for hh.

    tan⁡30∘=h20 m\tan 30^\circ = \frac{h}{20 \text{ m}} h=20 m×tan⁡30∘h = 20 \text{ m} \times \tan 30^\circ

    We know that tan⁡30∘=13≈0.577\tan 30^\circ = \frac{1}{\sqrt{3}} \approx 0.577.

    h=20 m×0.577h = 20 \text{ m} \times 0.577 h=11.54 mh = 11.54 \text{ m}
  12. Compare with the given options.

    The calculated value 11.54 m11.54 \text{ m} is closest to option C.

    The final answer is C. 11.5 m\boxed{\text{C. } 11.5 \text{ m}}.

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