Math Case Study Questions Class 10 on Height and Distance

Class 10 Math Height and Distance Case Study Questions | Trigonometry Case Studies Math Case Study Questions Class 10 on Height and Distance | Trigonometry Case Studies

Case Study 3

A tall street light pole stands vertically on the side of a straight road. On a sunny day, the pole casts a shadow on the road. Students of Class 10 decided to calculate the angle of elevation of the Sun by measuring the length of the pole and its shadow. They also verified trigonometric identities using the calculated values of sine, cosine, and tangent. This real-life context connects basic trigonometric ratios to a practical observation. It helps to demonstrate how identities like \(\sin^2\theta + \cos^2\theta = 1\) and \(1 + \tan^2\theta = \sec^2\theta\) can be applied to everyday situations involving right-angled triangles.

1. A pole \(12\) m high casts a shadow of length \(5\) m. The hypotenuse of the right triangle formed is:

  • A) 12 m
  • B) 13 m
  • C) 14 m
  • D) 15 m
Answer: B) 13 m
Solution: Hypotenuse \(= \sqrt{12^2 + 5^2} = \sqrt{144+25} = \sqrt{169} = 13\).

2. The sine of the angle of elevation of the Sun is:

  • A) \(\tfrac{5}{12}\)
  • B) \(\tfrac{12}{13}\)
  • C) \(\tfrac{12}{5}\)
  • D) \(\tfrac{13}{12}\)
Answer: B) \(\tfrac{12}{13}\)
Solution: \(\sin\theta = \dfrac{12}{13}\).

3. The cosine of the angle of elevation of the Sun is:

  • A) \(\tfrac{5}{12}\)
  • B) \(\tfrac{12}{13}\)
  • C) \(\tfrac{5}{13}\)
  • D) \(\tfrac{13}{5}\)
Answer: C) \(\tfrac{5}{13}\)
Solution: \(\cos\theta = \dfrac{5}{13}\).

4. Verify which identity holds for these values of sine and cosine:

  • A) \(\sin^2\theta + \cos^2\theta = 1\)
  • B) \(\sin\theta \cos\theta = 1\)
  • C) \(\tan^2\theta + 1 = \csc^2\theta\)
  • D) \(\cos^2\theta – \sin^2\theta = 1\)
Answer: A) \(\sin^2\theta + \cos^2\theta = 1\)
Solution: \(\left(\tfrac{12}{13}\right)^2 + \left(\tfrac{5}{13}\right)^2 = \tfrac{144}{169} + \tfrac{25}{169} = \tfrac{169}{169} = 1\).

5. If \(\tan\theta = \tfrac{12}{5}\), then \(\sec^2\theta\) equals:

  • A) \(\tfrac{144}{25}\)
  • B) \(\tfrac{169}{25}\)
  • C) \(\tfrac{25}{169}\)
  • D) \(\tfrac{13}{12}\)
Answer: B) \(\tfrac{169}{25}\)
Solution: \(1 + \tan^2\theta = \sec^2\theta\) \(= 1 + \left(\tfrac{12}{5}\right)^2 = 1 + \tfrac{144}{25} = \tfrac{169}{25}\).

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