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Maths-

General

Easy

Question

In the figure if TP and TQ are the two tangents to a circle with centre O so that ÐPOQ = 110º, then ÐPTQ is equal to

$∠ P T Q = 60^{\circ}$
$∠ P T Q = 70^{\circ}$
$∠ P T Q = 80^{\circ}$
$∠ P T Q = 90^{\circ}$

The correct answer is: $angle P T Q equals 70 to the power of ring operator end exponent$

We should find the angle from the given figure
In the above figure, OPTQ is a quadrilateral and ∠P and ∠Q are 90°
The sum of the interior angles of a quadrilateral is 360°.
Therefore, in OPTQ,
∠Q + ∠P + ∠POQ + ∠PTQ = 360°
90° + 90° + 110° + ∠PTQ = 360°
290° + ∠PTQ = 360°
∠PTQ = 360° - 290°
∠PTQ = 70°

Hence ∠PTQ = 70°

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In the figure, $angle A O B equals 90 to the power of ring operator end exponent$ , AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of DA.OB.

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Line $a x plus b y plus p equals 0$ makes angle $pi divided by 4$ with $x blank c o s blank alpha plus y blank s i n blank alpha equals p comma p space element of R to the power of straight prime$ If these lines and the line $x blank s i n blank alpha minus y blank c o s blank alpha equals 0$ are concurrent, then

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