Question

Find the angle subtended by a chord of length 12 units to the center of a circle having a radius of √72 units.

degrees

Answer 1

The chord with the circle is shown in the figure, with P as the center of the circle.


$PU=\sqrt{72}\text{units}$ (radius)

PR is perpendicular to the chord UV from the center.
$\therefore$ PR will bisect UV at R.
$\Rightarrow UR=\frac{UV}{2}=\frac{12}{2}=6\text{units}$

Considering the right triangle $△UPR,$
$U{P}^2=U{R}^2+P{R}^2$
$\Rightarrow P{R}^2=U{P}^2-U{R}^2=(\sqrt{72}{)}^2-6^2$
$\Rightarrow PR=\sqrt{72-36}=\sqrt{36}=6\text{units}$

$\therefore PR=UR$
Hence, the corresponding angles of the opposite sides are also equal.
$\Rightarrow \angle PUR=\angle UPR$
From the property of sum of interior angles, $\angle PUR=\angle UPR={45}^o$

Similarly, for $△PVR,\angle VPR={45}^o.$

$\angle UPV=\angle UPR+\angle VPR={45}^o+{45}^o={90}^o$

$\therefore$ The required angle is ${90}^o.$