Question

OT = 7 cm and OP = 25 cm, find the length of PT. If PT′ is another tangent to the circle, find the length of PT′ and ∠POT.

Answer 1

Answer:

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Answer 2

Concept:

The angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

According to the Pythagoras theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides of a triangle.

The length of two of tangents drawn from an external point to a circle are equal.

Also, Sin∅=$\frac{Opposite}{Hypotenuse}$

Given:

The values of OT= 7 cm and OP= 25 cm

PT and PT' are tangents from external point P.

To find:

We need to find the length of PT and ∠POT.

Solution:

Now, from the given below figure and since ∠PTO is 90°

Therefore ΔOPT is a right angled triangle.

$OT^{2}+PT^{2} =OP^{2}$

⇒$PT^{2}=OP^{2} -OT^{2}$

⇒$PT^{2}=25^{2} -7^{2}=625-49=576$

⇒$PT=\sqrt{576}=24 cm$

Now, since PT and PT' are tangents from a single external point P then,

Therefore,

PT=PT'=24 cm

Now, in ΔPOT

Let, ∠POT=∅

sin∅=$\frac{PT}{PO}$

⇒sin∅=$\frac{24}{25}$

⇒sin∅=0.96

⇒∅=$sin^{-1}$(0.96)

⇒∅=73.739795°

Therefore, PT'=24 cm and ∠POT=73.739795°