Question

A circular garden of radius 10m has a straight line fence between the two points on the boundary of the garden. The fence lies inside the garden arena. This fence separates the walking area which is a small region from the plants. The fence is at a distance of 6 m from the centre of the garden. Find the angle subtended by the fence with the center of the garden. It is given that cos(${53}^{\circ }$) = $\frac{3}{5}$.

Answer 1

${10}^2$ = ${OR}^2$ + ${RQ}^2$
${10}^2$ = $6^2$ +${RQ}^2$
$\Rightarrow$ ${RQ}^2$ = 64
$\Rightarrow$ RQ = 8m
$\therefore$ PQ = 2RQ = 16m (OR is perpendicular bisector of PQ) (1 mark)

Area of $\Delta$OPQ = $\frac{1}{2}$ × Base × Height
= $\frac{1}{2}$ × PQ × OR
= $\frac{1}{2}$ × 16 × 6
$=48m^2$ (1 mark)

Area of sector OPSQ = $\frac{\angle POQ}{{360}^{\circ }}$ × $\pi \times r^2$
In ORQ,
$cos(\angle$ROQ) = $\frac{adjacentside}{hypotenuse}$
= $\frac{OR}{OQ}$
= $\frac{6}{10}$
= $\frac{3}{5}$ (1 mark)
$\Rightarrow$ cos( $\angle$ROQ) = $\frac{3}{5}$
It is given that $cos({53}^{\circ }$) = $\frac{3}{5}$.
Hence $\angle$ROQ = ${53}^{\circ }$.
The angle subtended by the fence with the centre of the garden is $\angle$POQ
$⟹\angle$POQ = 2($\angle$ROQ) = ${106}^{\circ }$ (1 mark)