Question

14. If angle between two tangents drawn from
a point P to a circle of radius A and centre o
is 90°, then find the value of OP​

Answer 1

Answer:

Step-by-step explanation:

The length OP is a\sqrt{2}a

2

units.

Let the two tangents drawn from an external point P to the circle be PQ and PR.

Given,

O is center of the circle with radius a.

OQ = OR = a

∠QPR = 90°

OP is the angle bisector of ∠QPR

⇒ ∠OPQ = ∠OPR = 45°

We know that, the tangent drawn to a circle is perpendicular to radius.

⇒ ∠OQP = ∠ORP = 90°

Now, ΔOPQ is a right angled triangle

\begin{gathered}sin\alpha = \frac{opp}{hyp}\\sin45 = \frac{OQ}{OP} \\\frac{1}{\sqrt{2} }=\frac{a}{OP} \\text{\O}P = a\sqrt{2} units\end{gathered}

sinα=

hyp

opp

sin45=

OP

OQ

2

1

=

OP

a

text\OP=a

2

units