Question

In figure if TP and TQ are the two tangents to a circle with center O so that $\angle POQ = {110^0},$ the $\angle PTQ$ is equal to.

A.${60^0}$
B.${70^0}$
C.${80^0}$
D.${90^0}$

Answer 1

Hint- In order to solve this question, we will use the basic properties of angles and quadrilaterals. As we know that the sum of interior angles of a quadrilateral is 360 degree and the radius of the circle always makes an angle of 90 degree with the tangent to the circle. We will use this property to find the solution.

Complete step-by-step answer:
Given
$\angle POQ = {110^0}$
Also tangent at any point of circle is perpendicular to the radius through the point of contact
Here OP and OQ is the radius of the circle and PT and QT are the tangents to the circle.
Since OP and PT are perpendicular to each other
$\angle OPT = {90^0}$
Similarly, OQ and QT are perpendicular to each other
$\angle OQT = {90^0}$
In quadrilateral TPOQ
As we know that sum of interior angles of a quadrilateral is 360 degree
$\angle PTQ + \angle TPO + \angle POQ + \angle OQT = {360^0}$
Substituting the values of these angles
$
  {90^0} + {90^0} + {110^0} + \angle PTQ = {360^0} \\
  {290^0} + \angle PTQ = {360^0} \\
  \angle PTQ = {360^0} - {290^0} \\
  \angle PTQ = {70^0} \\
$
Therefore, the angel $\angle PTQ = {70^0}$
Hence, the correct option is B.
Additional Information- A tangent to a circle is a line which touches the circle at only one point. The tangent to the circle forms a right angle with the radius of the circle at the point of contact.

Note- In order to solve these types of questions, learn the basic properties of angles and also about the sum of interior angles of various shapes such as triangle, square, hexagon. Read the question carefully and extract all the data given in the question required to find the solution. Also remember, when you need to frame an equation and find the solution of the equations; to find the solutions the number of equations framed must be equal to the number of variables.