To solve both problems, we'll break them down step by step.
Problem 1: Height of the Cone
Given:
- Vertical angle \( \theta = 70^\circ \)
- Slant height \( l = 11 , \text{m} \)
We can find the height \( h \) of the cone using trigonometric relationships. The vertical angle is the angle between the slant height and the height of the cone. We can use the cosine of half the vertical angle to find the height.
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Find the angle from the vertical: \[ \text{Half of the vertical angle} , = \frac{70^\circ}{2} = 35^\circ \]
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Use the cosine relation: \[ \cos(35^\circ) = \frac{\text{adjacent side (height)}}{\text{hypotenuse (slant height)}} \] Rearranging gives: \[ h = l \cdot \cos(35^\circ) \] We need to calculate \( \cos(35^\circ) \): \[ \cos(35^\circ) \approx 0.8192 \quad (\text{using a calculator}) \]
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Calculate the height: \[ h = 11 \cdot 0.8192 \approx 9.0112 , \text{m} \]
Thus, the height of the cone is approximately 9.01 m.
Problem 2: Depth of Water in the Cylindrical Tank
Given:
- Volume of water \( V = 1760 , \text{litres} = 1760 , \text{dm}^3 = 1.76 , \text{m}^3 \) (since \( 1 , \text{litre} = 1 , \text{dm}^3 \))
- Diameter of the tank = 200 cm = 2 m (since \(1 , \text{m} = 100 , \text{cm}\))
- Radius \( r = \frac{diameter}{2} = 1 , \text{m} \)
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Formula for the volume of a cylinder: \[ V = \pi r^2 h \] We need to find \( h \): \[ h = \frac{V}{\pi r^2} \]
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Calculate \( \pi r^2 \): \[ r^2 = 1^2 = 1 , \text{m}^2 \] \[ \pi r^2 = \frac{22}{7} \times 1 = \frac{22}{7} , \text{m}^2 \]
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Calculate the depth of water (height): \[ h = \frac{1.76}{\frac{22}{7}} = 1.76 \times \frac{7}{22} \] \[ h = \frac{12.32}{22} \approx 0.56 , \text{m} \]
Thus, the depth of water in the cylindrical tank is approximately 0.56 m (or 56 cm).
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