(a) In the diagram, < RTS = 28°, < VRM = 46°, MQ is a tangent to the circle VRSTU at the point R. Find < VUS. (b) A cylinder tin, 7cm high, is closed at one...

Answer Details

(a) Finding \(\angle VUS\).

The points V, R, S, T, U lie on a circle, and line MRQ is a tangent at R. Given \(\angle RTS = 28^\circ\) and \(\angle VRM = 46^\circ\).

Step 1 - use the tangent-chord (alternate segment) property on chord RV. The angle between the tangent RM and the chord RV equals the inscribed angle in the alternate segment standing on RV. Hence the arc VR (the intercepted arc) satisfies

\[\angle VRM = \tfrac{1}{2}(\text{arc } VR) \;\Rightarrow\; \text{arc } VR = 2 \times 46^\circ = 92^\circ.\]

Step 2 - use the inscribed angle on chord RS. \(\angle RTS = 28^\circ\) is the angle at the circumference standing on chord RS, so

\[\text{arc } RS = 2 \times 28^\circ = 56^\circ.\]

Step 3 - find \(\angle VUS\). \(\angle VUS\) is the inscribed angle at U standing on chord VS. It equals half the arc VS that does not contain U; that arc runs from V through R to S:

\[\text{arc } VRS = \text{arc } VR + \text{arc } RS = 92^\circ + 56^\circ = 148^\circ.\]

\[\angle VUS = \tfrac{1}{2}(148^\circ) = 74^\circ.\]

\[\boxed{\angle VUS = 74^\circ.}\]

(b) Radius of the cylindrical tin.

The tin is closed at one end only, so its total surface area is the curved surface plus one circular end:

\[A = 2\pi r h + \pi r^2 = 462,\qquad h = 7,\ \pi = \tfrac{22}{7}.\]

Substitute:

\[2 \times \tfrac{22}{7} \times r \times 7 + \tfrac{22}{7} r^2 = 462.\]

\[44r + \tfrac{22}{7}r^2 = 462.\]

Multiply every term by 7:

\[308r + 22r^2 = 3234.\]

Divide through by 22:

\[14r + r^2 = 147 \;\Rightarrow\; r^2 + 14r - 147 = 0.\]

Solve:

\[r = \frac{-14 \pm \sqrt{14^2 + 4(147)}}{2} = \frac{-14 \pm \sqrt{196 + 588}}{2} = \frac{-14 \pm \sqrt{784}}{2} = \frac{-14 \pm 28}{2}.\]

Taking the positive root: \(r = \dfrac{14}{2} = 7\text{ cm}.\)

Check: \(2(\tfrac{22}{7})(7)(7) + (\tfrac{22}{7})(7^2) = 308 + 154 = 462\text{ cm}^2.\) \(\;\checkmark\)

\[\boxed{r = 7\text{ cm}.}\]