In the given figure, PT is a tangent to the circle with centre O and radius r. If $\angle POT = 45^\circ$, then the length of OP is :
Show Hint
In a right-angled triangle with an angle of $45^\circ$, the other acute angle is also $45^\circ$ ($90^\circ - 45^\circ = 45^\circ$).
This makes it an isosceles right-angled triangle where the two perpendicular sides are equal: $OT = PT = r$.
Using the Pythagorean theorem directly:
\[ OP = \sqrt{OT^2 + PT^2} = \sqrt{r^2 + r^2} = \sqrt{2r^2} = r\sqrt{2} \]
This visual and geometric approach bypasses trigonometry completely and is highly intuitive.
Step 1: Find the third angle of the triangle instead of using the cosine ratio directly.
In right triangle $OTP$, the tangent $PT$ meets the radius $OT$ at a right angle, so $\angle OTP = 90^\circ$. We are given $\angle POT = 45^\circ$. Since the angles of a triangle add up to $180^\circ$:
\[ \angle OPT = 180^\circ - 90^\circ - 45^\circ = 45^\circ \]
Step 2: Notice the triangle is isosceles right-angled, and use that to compare the two legs.
Since $\angle POT = \angle OPT = 45^\circ$, the sides opposite these equal angles must also be equal. The side opposite $\angle POT$ is $PT$, and the side opposite $\angle OPT$ is $OT$. So:
\[ PT = OT = r \]
Step 3: Use Pythagoras' theorem on the hypotenuse OP, instead of the cosine ratio.
Since $OP$ is the hypotenuse of right triangle $OTP$:
\[ OP^2 = OT^2 + PT^2 = r^2 + r^2 = 2r^2 \]
\[ OP = \sqrt{2r^2} = r\sqrt{2} \]
Final Answer:
The length of $OP$ is $r\sqrt{2}$, which matches Option (A).
\[ \boxed{OP = r\sqrt{2}} \]
