5 Marks Questions

Question 15 Marks

Find the distance from the eye at which a coin of 2cm diameter should be held so as to conceal the full moon whose angular diameter is 31'.

Answer

Let, r be the distance at which coin in placed. So that it completely conceals the full moon.
Let, E be the eye of the observer.
Now,
$\theta=31'= \Big(\frac{31}{60}\Big)^{\circ}$
$=\frac{31}{60}\times\Big(\frac{\pi}{180}\Big)^{\circ}$
Also,
$\theta= \frac{\text{arc}}{\text{radius}}$
$=\frac{31\pi}{60\times180}=\frac{0.02}{\text{r}}$
$=2.217\text{m}$
The coin should be placed at a distance of 2.217m from the eye.

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Question 25 Marks

A wheel makes 360 revolutions per minute. Through how many radians does it turn in 1 second?

Answer

Since A wheel makes 360 revolution in 1 minutes.
wheel will make $\frac{360}{60}$ revolution in 1 secons
That is, 6 revolution in 1 second
Now,
In one revolution the wheel makes 360° angle.
In 6 revolution the wheel makes 360° angle = 2160°
$1^{\circ}=\Big(\frac{\pi}{180}\Big)^{\text{c}}$
$2160^{\circ}=\Big(\frac{2160}{180}\times\pi\Big)^{\circ}$
$=12\pi$

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Question 35 Marks

If the arcs of the same length in two circles subtend angles $65^\circ$ and $110^\circ $at the centre, find the ratio of their radii.

Answer

Let, $C_1 \& C_2$ are two cirdes with same Arc length.
That is $AB = AB$
Let, $\theta_{1}$ and $\theta_{2}$ are two angle by arc $AB$ and $CD$ on respective circle.
Let, $OA = OB = r$
Also,
$\theta_{1}=65^{\circ}= \Big(\frac{65\pi}{180}\Big)^{\text{c}}$
$\theta_{2}=110^{\circ}= \Big(\frac{110\pi}{180}\Big)^{\text{c}}$
we know,
$\theta= \frac{\text{arc}}{\text{radius}}$
For $C_1$
$\theta_{1}=\frac{\text{AB}}{\text{r}}$
$\Rightarrow \theta_{1}=\frac{\text{l}}{\text{r}}$
$\Rightarrow\text{r}= \frac{1}{\theta_{1}}\ ...(\text{i})$
For $C_2$
$\theta_{2}=\frac{\text{AB}}{\text{R}}$
$\Rightarrow \theta_{2}=\frac{\text{l}}{\text{R}}$
$\Rightarrow\text{R}= \frac{1}{\theta_{2}}\ ...(\text{ii})$
From (i) and (ii),
$\frac{\text{r}}{\text{R}}= \frac{\frac{1}{\theta_{1}}}{\frac{1}{\theta_{2}}}=\frac{\theta_{2}}{\theta_{1}}$
$=\frac{\frac{110\pi}{180}}{\frac{65\pi}{180}}=\frac{22}{13}$
$\therefore \ \text{r}:\text{R}= 22:13$

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Question 45 Marks

A railway train is travelling on a circular curve of 1500 metres radius at the rate of 66km/ hr. Through what angle has it turned in 10 seconds?

Answer

We have,
In circle track
OA = OB = r = 150m
$\angle\text{AOB}= \theta$
Speed of train = 66km/ hr
$=\frac{66\times1000}{60\times60} $
$=\frac{110}{6}$
Train will travel in 10 sec $=\frac{110}{6}\times10=\frac{1100}{6}$
$\text{AB}= \frac{11000}{6}\text{m}$
Thus,
$\theta= \frac{\text{arc}}{\text{radius}}$
$\Rightarrow \frac{11000}{6\times1500}= \frac{11}{90}\ \text{radian}$
The train will turn by angle in 10 sec.

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Question 55 Marks

Find the diameter of the sun in km supposing that it subtends an angle of 32' at the eye of an observer. Given that the distance of the sun is $91 \times 10^6km.$

Answer

Let, E be the eye of the observer and be the sum.
Now,
$\angle\text{AOB}= \theta= 32'$
$=\Big(\frac{32}{60}\Big)^{\circ}$
$=\Big(\frac{32}{60}\times\frac{\pi}{180}\Big)^{\text{c}}$
$=\frac{\text{AB}}{91\times10^{6}}\text{km}$
$\text{AB}= \frac{91\times10^{6}\times32\times\pi}{60\times180}$
$=8.474074\times10^{5}\text{km}$
$=84707.4\text{km}$

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Question 65 Marks

Find the angle in radians through which a pendulum swings if its length is $75\ cm$ and the tip describes an arc of length.

$10\ cm$
$15\ cm$
$21\ cm$

Answer

we have,

$OA =$ length of pendulum $= 75\ cm$
$= 0.75m$
Also,
$\theta= \frac{\text{arc}}{\text{radius}}\ ...(\text{i})$
$\Rightarrow\theta= \frac{0.1}{0.75}=\Big(\frac{2}{15}\Big)^{\text{c}}$
$\theta=\frac{2}{15}\ \text{radian}$

$OA = 75\ cm$

$= 0.75\ cm$
$AB = 15\ cm$
$= 0.15m$
From $(i),$
$\theta- \frac{0.15}{0.75}=\frac{1}{5}$

$OA = 75\ cm$

$= 0.75m$
$AB = 21\ cm$
$= 0.21m$
From $(i)$
$\theta= \frac{0.21}{0.75}=\frac{7}{5}$

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Question 75 Marks

The angle of a quadrilateral are in A.P. and the greatest angle is 120°. Express the angles in radians.

Answer

Let the angle in degrees be a - 3d, a - d, a + 3d
Then,
Sum of the angle = 360°
⇒ 4a = 360°
a = 90°
Also,
Greatest angle = 120°
⇒ a + 3d = 120°
⇒ 3d = 30°
⇒ d = 10°
Hence, angle in degrees
60°, 80°, 100°, 120° and in redians,
We know that,
$1^{\circ}=\Big(\frac{\pi}{180}\Big)^{\text{c}}$
$\therefore 60\times\frac{\pi}{180}=\frac{\pi}{3},80\times\frac{\pi}{180}$
$=\frac{4\pi}{9}$
$100\times\frac{\pi}{180}=\frac{5\pi}{9},120\times\frac{\pi}{180}$
$=\frac{2\pi}{3}$
$\therefore \frac{\pi}{3},\frac{4\pi}{9},\frac{5\pi}{9},\frac{2\pi}{3}$

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Question 85 Marks

The number of sides of two regular polygons are as 5 : 4 and the difference between their angles is 9°. Find the number of sides of the polygons.

Answer

Let the number of sides in the first polygon be the number of sides in the second polygon.We know that each angle of regular polygon is $\Big(\frac{2\text{n}-4}{\text{n}}\Big)$ right angles.
Now,
According to the quation,
$\frac{\text{n}}{\text{m}}=\frac{5}{4}$
$\Rightarrow \frac{5\text{m}}{4}=\text{n}\ ...(\text{i})$
Also,
$\Big(\frac{2\text{n}-4}{\text{n}}\Big)\times90^{\circ}-\Big(\frac{2\text{m}-4}{\text{m}}\Big)\times90^{\circ}=9^{\circ}$
$\Rightarrow \frac{(2\text{n}-4)\text{m}-(2\text{m}-4)\text{n}}{\text{mn}}=\Big(\frac{1}{10}\Big)^{\circ}\ ...(\text{ii})$
From (i) and (ii),
$\frac{\big(2\times\frac{5}{4}\text{m}-4\big)\text{m}-\big(2\text{m}-4\big)\frac{5}{4}\text{m}}{\frac{5}{4}\text{m}^{2}}=\frac{1}{10}$
$\Rightarrow \frac{(10\text{m}-16)-(10\text{m}-20)}{5\text{m}}=\frac{1}{10}$
$\Rightarrow \frac{4}{\text{m}}=\frac{1}{2}$
$\Rightarrow \text{m}=8$

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Question 95 Marks

The angles of a triangle are in A.P. such that the greatest is 5 times the least. Find the angles in radians.

Answer

Let A, B & C are the angles of triangle ABC
According to the quation,
Also,
A, B & C are in A.P.
Let A = a - d, B = a & C = a + d
So, A + B + C = 180°
⇒ a - d + a + a + d = 180°
⇒ 3a = 180°
⇒ a = 60° ...(i)
Also,
Greatest angle in 5 times the least
a + d = 5(a - d)
⇒ 4a = 6d
$\Rightarrow \frac{2}{3}\text{a}$
$\Rightarrow \text{d}=\frac{2}{3}\times60=40^{\circ}\ ...(\text{ii})$
$\therefore 1^{\circ}=\Big(\frac{\pi}{180^{\circ}}\Big)\ \text{radians}$
$\text{A}=20\times\frac{\pi}{180}=\frac{\pi}{9}$
$\text{B}=60\times\frac{\pi}{180}=\frac{\pi}{3}$
$\text{C}=100\times\frac{\pi}{180}=\frac{5\pi}{3}$
Thus,
$\text{A}=\frac{\pi}{9},\text{B}=\frac{\pi}{3},\text{C}=\frac{5\pi}{9}$

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