(a) In a right-angled triangle, sin X = \(\frac{3}{5}\). Evaluate, leaving the answer as a fraction, 5 (cosX)\(^2\) – 3. (b) The base of a pyramid, 12 cm h...
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(a) In a right-angled triangle, sin X = \(\frac{3}{5}\). Evaluate, leaving the answer as a fraction, 5 (cosX)\(^2\) – 3.
(b) The base of a pyramid, 12 cm high, is a rectangle with dimensions 42 cm by 11 cm. if the pyramid is filled with water and emptied into a conical container of equal height and volume, calculate, leaving the answer in surd form (radicals), the base radius of the container. [Take π=\(\frac{22}{7}\)]
Answer Details
(a) \(\sin X = \tfrac{3}{5} \Rightarrow \cos X = \tfrac{4}{5}\) (a 3-4-5 triangle).
\[5\cos^2 X - 3 = 5\left(\tfrac{16}{25}\right) - 3 = \tfrac{16}{5} - 3 = \tfrac{1}{5}\](b) Volume of the pyramid \(= \tfrac{1}{3} \times (42 \times 11) \times 12 = \tfrac{1}{3} \times 462 \times 12 = 1848 \text{ cm}^3\).
The cone has the same height (12 cm) and the same volume (1848):
\[\tfrac{1}{3} \times \tfrac{22}{7} \times r^2 \times 12 = 1848\] \[\tfrac{88}{7}r^2 = 1848 \Rightarrow r^2 = \frac{1848 \times 7}{88} = 147\] \[r = \sqrt{147} = 7\sqrt{3} \text{ cm}\]Download The App On Google Playstore
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