Question

The radius of a circle is given as 15 cm and chord AB subtends an angle of ${131}^{\circ }$ at the centre C of the circle. Using trigonometry, calculate :

(i) the length of AB;

(ii) the distance of AB from the centre C.

Answer 1

Solution:-
Consider the circle with radius equal to 15 cm with center 'O' and AB be the chord which subtends an angle of 131° at the center of the circle.
As we know that perpendicular from the center of the chord bisects the chord.
Let AB = x cm
⇒ AC = $\frac{x}{2}$ cm
$\angle AOB={131}^0$
$\therefore \angle AOC=\frac{{131}^0}{2}={65.5}^0$
In$△AOC$,
$sin\left({65.5}^0\right)=\frac{x}{15\ast 2}$
$=0.909\times 30(Assin{65.5}^0=.909)$
$\to x=27.27cm$
So, the length of AB is 27.27 cm
$cos{65.5}^0=\frac{OC}{15}$
⇒ OC = .414 × 15
OC = 6.21 cm
Hence, the length of chord AB is 27.27 cm and distance between the center and the chord is 6.21 cm.