1 Marks Question

Question 11 Mark

From an external point $P$, tangents $P A$ and $P B$ are drawn to a circle with centre $O$. If $\angle P A B=50^{\circ},$ then find $\angle A O B$.

Answer

It is given that $P A$ and $P B$ are tangents to the given circle.
$\therefore \angle P A O=90^{\circ} \ ($Radius is perpendicular to the tangent at the point of contact.$)$
Also, $\angle P A B=50^{\circ}$
$\therefore \angle O A B=\angle P A O-\angle P A B$
$=90^{\circ}-50^{\circ}=40^{\circ}$
In $\triangle O A B, O B=O A \ \ ($Radii of the circle$)$
$\therefore \angle O A B=\angle O B A=40^{\circ} \ ($Angles opposite to equal sides are equal.$)$
Now, using the angle sum property of triangles,
$\angle A O B+\angle O A B+\angle O B A=180^{\circ}$
$\angle A O B=180^{\circ}-40^{\circ}-40^{\circ}=100^{\circ}$
Hence, $\angle A O B=100^{\circ}$

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Question 21 Mark

If the angle between two tangents drawn from an external point $P$ to a circle of radius a and centre $O,$ is $60^{\circ},$ then find the length of $O P$.

Answer

Along with center $O, P A$ and $P B$ are two tangents drawn to the circle
$\angle A P B=60^{\circ}$
In $\triangle O P B$ and $\triangle O P A$
The radii of the circle is equal $( O B=O A=a ) \angle O B P=\angle O A P=90^{\circ} \ldots$.
At point of contact tangents are perpendicular to radius
$B P=P A \ldots...$
From an external point to the circle lengths of tangents drawn are equal
So, $\triangle O P B \cong \triangle O P A \ (\text{SAS}$ congruence$)$
$\therefore \angle O P B=\angle O P A=30^{\circ} \ ($By $\text{CPCT})$
In $\triangle O P B$
$\sin 30^{\circ}=\frac{O B}{O P}$
$\Rightarrow \frac{1}{2}=\frac{a}{O P} $
$\Rightarrow O P=2 a$
Therefore, the length of $O P=2 a$

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Question 31 Mark

In Fig. $, AB$ is the diameter of a circle with centre $O$ and $A T$ is a tangent. If $\angle A O Q=58^{\circ},$ find $\angle A T Q$.

Answer

It is given that $\angle A O Q=58^{\circ}$.
Therefore $\angle A B Q=\frac{1}{2} \times \angle A O Q $
$ ($Since angle subtended by an arc at the centre of the circle is twice the angle subtended by it at any point on the remaining part of the circle$).$
$\Rightarrow \angle A B Q=\frac{1}{2} \times 58^{\circ}$
$ \Rightarrow \angle A B Q=29^{\circ}$
Now we know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore $\angle B A T=90^{\circ}$
In $\triangle A B T$,
$\angle A B T+\angle B A T+\angle B T A=90^{\circ} $
$ ($Since the sum of all the angles in a triangle is $180^{\circ} )$
$\Rightarrow 29^{\circ}+90^{\circ}+\angle B T A=180^{\circ}$
$\Rightarrow \angle B T A=180^{\circ}-119^{\circ}$
$\Rightarrow \angle B T A=61^{\circ}$
Hence, $\angle A T Q=61^{\circ}$

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Question 41 Mark

In given Fig., $P A$ and $P B$ are tangents to the circle with centre 0 such that $\angle A P B=50^{\circ}$. Write the measure of $\angle O A B$.

Answer

Here $P A$ and $P B$ are the two tangents.
We know that the tangents drawn from the external point are equal.
$\therefore P A=P B$
The $\triangle P A B$ is an isosceles triangle with $P A=P B$
Therefore $\angle P A B=\triangle P B A ...(i)$
$($Angles opposite to equal sides$)$
In $\triangle P A B$,
$\angle A P B+\angle P A B+\angle P B A=180^{\circ}$
$\Rightarrow 50^{\circ}+2 \angle P A B=180^{\circ} \ \ ($ Using $(i))$
$2 \angle P A B=180^{\circ}-50^{\circ}$
$\Rightarrow \angle P A B=\frac{130^{\circ}}{2}$
$\Rightarrow \angle P A B=65^{\circ}$
Now we know that the radius is perpendicular to the tangent at the point of contact.
$\therefore \angle O A B+\angle P A B=90^{\circ}$
$\Rightarrow \angle O A B+65^{\circ}=90^{\circ}\ \ ($Using $(ii))$
$\Rightarrow \angle O A B=90^{\circ}-65^{\circ}$
$\Rightarrow \angle O A B=25^{\circ}$
Hence $\angle O A B=25^{\circ}$

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Question 51 Mark

In given Figure, the length $PB =$ _____________ cm .

Answer

$
\begin{array}{rlrl}
OA^2 =OP^2+AP^2 \\
\Rightarrow 5^2 =3^2+AP^2 \\
\Rightarrow AP^2 =25-9 \\
\Rightarrow AP^2 =16 \\
\Rightarrow AP =4 cm \\
\Rightarrow AB =2 AP \\
\Rightarrow AB =2 \times 4 \\
AB =8 cm
\end{array}
$

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Question 61 Mark

A line intersecting a circle in two points is called a ______________ .

Answer

Secant

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