(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=\(\frac{22}{7}\)]
(a) By the use of Pythagoras theorem, the adjacent side of the right – angled triangle is 4.
Then, cos(X = \(\frac{4}{5}\)) and substituting to get 5(cos X)\(^2\) – 3
= 5(\(\frac{4}{5}\))\(^2\) - 3
= \(\frac{16}{5}\) - 3
= \(\frac{(16 - 15)}{5}\) = \(\frac{1}{5}\)
(b) find Volume of pyramid = \(\frac{1}{3}\) × 42 × 11 × 12 = 1,848 (cm)\(^2\)
Volume of conical container = \(\frac{1}{3}\) × \(\frac{22}{7}\) x r\(^2\) × 12 = 1,848
and simplifying to get r\(^2\) = 147 so that r =7√3 cm as required.
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}