2 Marks Questions

Question 12 Marks

Prove that: $\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\tan \theta+\cot \theta$

Answer

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\tan \theta+\cot \theta$
$\text{L.H.S.} =\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}$
$=\frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}$
$=\frac{\sin ^2 \theta}{\cos \theta(\sin \theta-\cos \theta)}-\frac{\cos ^2 \theta}{\sin \theta(\sin \theta-\cos \theta)}$
$=\frac{\sin ^3 \theta-\cos ^3 \theta}{\sin \theta \cos \theta(\sin \theta-\cos \theta)}$
$=\frac{(\sin \theta-\cos \theta)\left(\sin ^2 \theta+\cos ^2 \theta+\sin \theta \cos \theta\right)}{\sin \theta \cos \theta(\sin \theta-\cos \theta)}\left[\because a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right.$
$=\frac{\sin ^2 \theta}{\sin \theta \cos \theta}+\frac{\cos ^2 \theta}{\sin \theta \cos \theta}+\frac{\sin \theta \cos \theta}{\sin \theta \cos \theta}$
$=\tan \theta+\cot \theta+1$
$=1+\tan \theta+\cot \theta$
$=\ce{RHS}$
Hence proved.

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Question 22 Marks

A horse is tethered to one corner of a rectangular field of dimensions $70 m \times 52 m$, by a rope of length $21 m .$ How much area of the field can it graze?

Answer

Shaded portion indicates the area which the horse can graze. Clearly, shaded area is the area of a quadrant of a circle of radius $r - 21 m.$

$\therefore$ Required Area $=\frac{1}{4} \pi r^2$
$\therefore$ Required Area $=\left\{\frac{1}{4} \times \frac{22}{7} \times(21)^2\right\} \ cm ^2$
$=\frac{693}{2} \ cm^2$
$=346.5 \ cm^2$

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Question 32 Marks

Find the length of the arc of a circle of diameter $42 \ cm$ which subtends an angle of $60^{\circ}$ at the centre.

Answer

$\text { Diameter of a circle }=42 \ cm$
$\Rightarrow \text { Radius of a circle }= r =\frac{42}{2}$
$=21 \ cm$
$\text { Central angle }=\theta=60^{\circ}$
$\therefore \text { Length of the arc }=\frac{2 \pi r \theta}{360}$
$=\frac{2 \times \frac{22}{7} \times 21 \times 60^{\circ}}{360^{\circ}} \ cm$
$=22 \ cm$

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Question 42 Marks

Show that: $\tan ^4 \theta+\tan ^2 \theta=\sec ^4 \theta-\sec ^2 \theta$ for $0^{\circ} \leq \theta \geq 90^{\circ}$

Answer

$\text { L.H.S }=\tan ^4 \theta+\tan ^2 \theta$
$=\tan ^2 \theta\left(\tan ^2 \theta+1\right)$
$=\tan ^2 \theta \sec ^2 \theta$
$=\left(\sec ^2 \theta-1\right) \sec ^2 \theta\left[\because \tan ^2 \theta=\sec ^2 \theta-1\right]$
$=\sec ^4 \theta-\sec ^2 \theta$
$=\text { R.H.S. }$

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Question 52 Marks

Two concentric circles with centre O are of radii 3 cm and 5 cm. Find the length of chord AB of the larger circle which touches the smaller circle at P.

Answer

Join OA and OP
$OP \perp AB$ (radius $\perp$ tangent at the point of contact)

OP is the radius of smaller circle and $A B$ is tangent at $P$.
AB is chord of larger circle and $OP \perp AB$
$\therefore AP = PB (\perp$ from centre bisects the chord $)$
In right $\triangle AOP , AP ^2= OA ^2- OP ^2$
$=(5)^2-(3)^2=16$
$AP =4 cm= PB$
$\therefore AB =8 cm$

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Question 62 Marks

In the given figure $ XY \| BC$. Find the length of $XY$.

Answer

Given $ X Y \| B C$
$A X=1 \ cm, X B=3 \ cm , $ and $ B C=6 \ cm$
$A B=A X+X B$
$=1+3=4 \ cm$
In $\triangle AXY$ and $\Delta ABC$
$\angle A=\angle A \ [$Common$]$
$\angle A X Y=\angle A B C \ [$Corresponding angles$]$
Then, $\triangle AXY \sim \triangle ABC \ [$By $AA$ similarity$]$
$\therefore \frac{A X}{A B}=\frac{X Y}{B C}\ [$Corresponding parts of similar $\triangle$ are proportional$]$
$\Rightarrow \frac{1}{4}=\frac{X Y}{6}$
$\Rightarrow X Y=\frac{6}{4}=1.5 \ cm$

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Question 72 Marks

Find the largest number which divides $320$ and $457$ leaving remainder $5$ and $7$ respectively.

Answer

The given numbers are $320$ and $457$
Now as $5$ and $7$ are remainders on division of $320$ and $457$ by said number
On subtracting the reminders $5$ and $7$ from $320$ and $457$ respectively we get:
$320-5=315$
$457-7=450$
The prime factorizations of $315$ and $405$ are
$315=3 \times 3 \times 5 \times 7$
$=3^2 \times 5 \times 7$
$450=2 \times 3 \times 3 \times 5 \times 5$
$=2 \times 3^2 \times 5^2$
$\therefore \text{H.C.F.}$ of $315$ and $450=3^2 \times 5=9 \times 5=45$
Hence the said number $=45$

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Maths 2 Marks Questions

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