$\text{From an external point}A\text{, tangents}\mathrm{AP}\text{and}\mathrm{AQ}\text{are drawn to a circle with centre}O\text{. If}\angle \mathrm{APQ}=40°\text{, then find}\angle \mathrm{POQ}.$

The correct option is 80°.

Step 1: Understand the properties of tangents from an external point.
When two tangents are drawn to a circle from an external point A, their lengths are equal.
$\mathrm{AP}=\mathrm{AQ}$

Step 2: Determine the angles in triangle APQ.
Since $\mathrm{AP}=\mathrm{AQ}$, triangle APQ is an isosceles triangle with equal angles opposite to the equal sides:
$\angle \mathrm{AQP}=\angle \mathrm{APQ}=40°$

Step 3: Calculate angle PAQ.
The sum of interior angles in triangle APQ is equal to 180°:
$\angle \mathrm{PAQ}+\angle \mathrm{APQ}+\angle \mathrm{AQP}=180°$
Substitute the given values:
$\angle \mathrm{PAQ}+40°+40°=180°$
$\angle \mathrm{PAQ}+80°=180°$
$\angle \mathrm{PAQ}=180°-80°=100°$

Step 4: Find angle POQ.
The tangent at any point of a circle is perpendicular to the radius through the point of contact:
$\angle \mathrm{APO}=90°$
$\angle \mathrm{AQO}=90°$

In quadrilateral APOQ, the sum of all interior angles is 360°:
$\angle \mathrm{PAQ}+\angle \mathrm{APO}+\angle \mathrm{POQ}+\angle \mathrm{AQO}=360°$
Substitute the known angle measures into the equation:
$100°+90°+\angle \mathrm{POQ}+90°=360°$
$\angle \mathrm{POQ}+280°=360°$
$\angle \mathrm{POQ}=360°-280°=80°$

Therefore, the value of $\angle \mathrm{POQ}$ is 80°.