If tangents \[PA\] and \[PB\] from a point \[P\] to a circle with centre \[O\] are inclined to each other at angle of \[80^\circ \], then \[\angle POA\] is equal to
A. \[50^\circ \]
B. \[60^\circ \]
C. \[70^\circ \]
D. \[80^\circ \]
Answer
568.9k+ views
Hint: In this problem, we have to find which is the correct option for \[\angle POA\].
Using the given data from the question we can convert it into a diagram for our understanding. Then by using the total angle of a triangle we are going to solve the problem.
SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent. ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent.
Complete step-by-step answer:
It is given that the tangents \[PA\] and \[PB\] from a point \[P\] to a circle with centre \[O\] are inclined to each other at an angle of \[80^\circ \].
We need to find out the measure of \[\angle POA\].
From,\[\Delta PAO\] \[\&\] \[\Delta PBO\]
\[PA = PB\]
Since tangents from external points are equal.
\[OA\] and \[OB\] are the radius of the circle,
Thus, \[OA = OB\]
And \[OP\] is common.
Therefore, \[\Delta PAO \cong PBO\](By SSS congruence)
Then,\[\angle OPA = \angle OPB\]
It is given that, \[\angle APB = 80^\circ \]
So we get, \[\angle OPA = \angle OPB = \dfrac{{80^\circ }}{2} = 40^\circ \]
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact.
\[\therefore \angle OAP = 90^\circ \]
Now from \[\Delta OAP\],
\[ \Rightarrow \angle OAP + \angle OPA + \angle POA = 180^\circ \]
Substituting the angle values of \[\angle OAP = 90^\circ \] and \[\angle OPA\]\[ = 40^\circ \]
\[ \Rightarrow 90^\circ + 40^\circ + \angle POA = 180^\circ \]
Solve the above equation for \[\angle POA\] we get,
\[ \Rightarrow \angle POA = 180^\circ - 90^\circ - 40^\circ \]
\[ \Rightarrow \angle POA = 50^\circ \]
Hence we get, \[\angle POA = 50^\circ \]
So, the correct answer is “Option A”.
Note: We have used here, the sum of the angles of a triangle is \[180^\circ \] .
Tangent: A straight line or plane that touches a curve or curved surface at a point, but if extended does not cross it at that point.
Property: The tangents from the external point are equal.
Using the given data from the question we can convert it into a diagram for our understanding. Then by using the total angle of a triangle we are going to solve the problem.
SSS (Side-Side-Side): If three pairs of sides of two triangles are equal in length, then the triangles are congruent. ASA (Angle-Side-Angle): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent.
Complete step-by-step answer:
It is given that the tangents \[PA\] and \[PB\] from a point \[P\] to a circle with centre \[O\] are inclined to each other at an angle of \[80^\circ \].
We need to find out the measure of \[\angle POA\].
From,\[\Delta PAO\] \[\&\] \[\Delta PBO\]
\[PA = PB\]
Since tangents from external points are equal.
\[OA\] and \[OB\] are the radius of the circle,
Thus, \[OA = OB\]
And \[OP\] is common.
Therefore, \[\Delta PAO \cong PBO\](By SSS congruence)
Then,\[\angle OPA = \angle OPB\]
It is given that, \[\angle APB = 80^\circ \]
So we get, \[\angle OPA = \angle OPB = \dfrac{{80^\circ }}{2} = 40^\circ \]
Since the tangent at any point of a circle is perpendicular to the radius through the point of contact.
\[\therefore \angle OAP = 90^\circ \]
Now from \[\Delta OAP\],
\[ \Rightarrow \angle OAP + \angle OPA + \angle POA = 180^\circ \]
Substituting the angle values of \[\angle OAP = 90^\circ \] and \[\angle OPA\]\[ = 40^\circ \]
\[ \Rightarrow 90^\circ + 40^\circ + \angle POA = 180^\circ \]
Solve the above equation for \[\angle POA\] we get,
\[ \Rightarrow \angle POA = 180^\circ - 90^\circ - 40^\circ \]
\[ \Rightarrow \angle POA = 50^\circ \]
Hence we get, \[\angle POA = 50^\circ \]
So, the correct answer is “Option A”.
Note: We have used here, the sum of the angles of a triangle is \[180^\circ \] .
Tangent: A straight line or plane that touches a curve or curved surface at a point, but if extended does not cross it at that point.
Property: The tangents from the external point are equal.
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