3 Marks Question

Question 13 Marks

Find the degree measure of the angle subtended at the centre of a circle of radius 100cm by an arc of length 22cm.

Answer

Let, AB = arc AB = 22cm
OA = OB = r = 100cm
Let $\theta$ bet the angle by arc AB at centre O.
$\theta=\frac{\text{arc}}{\text{radius}}$
$\theta=\frac{22}{100}\ \text{radius}$
$\theta= \Big(\frac{22}{100}\times\frac{180}{\pi}\Big)^{\circ}$
$=12.6^{\circ}$
$=12^{\circ}36'$

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Question 23 Marks

The radius of a circle is 30cm. Find the length of an arc of this circle, if the length of the chord of the arc is 30cm.

Answer

We have,
AO = OB
= raius of circle
= 30cm
= 0.3
Now,
$\triangle\text{AOB}$ in quailteral triangle as
OA = OB = AB
= 30
$\theta= \frac{\text{arc}}{\text{radius}}$
$\Rightarrow \frac{\pi}{3}=\frac{\text{l}}{0.3}$
$\Rightarrow \text{l} = \frac{0.3}{3}\pi= 0.1\pi\ \text{m} $

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Question 33 Marks

The angle in one regular polygon is to that in another as 3 : 2 and the number of sides in first is twice that in the second. Determine the number of sides of two polygons.

Answer

Let n & m be the number of sides in two regular polygon respectivelt.
We know that each angle of n sides regular is $\frac{(2\text{n}-4)}{\text{n}}$ right angles.
Now,
According to the quation,
$\frac{\Big(\frac{2\text{n}-4}{\text{n}}\Big)\times90^{\circ}}{\Big(\frac{2\text{m}-4}{\text{m}}\Big)\times90^{\circ}}=\frac{3}{2}$
$\Rightarrow \frac{(2\text{n}-4)\text{m}}{(2\text{m}-4)}=\frac{3}{2}\ ...(\text{i})$
Also,
$\text{n}=2\text{m}\ ...(\text{ii})$
Put (ii) in (i), we get
$\frac{(4\text{m}-4)\text{m}}{(2\text{m}-4)2\text{m}}=\frac{3}{2}$
$\Rightarrow 4\text{m}-4=6\text{m}-12$
$\Rightarrow 2\text{m}=8$
$\Rightarrow \text{m}=4$
From (ii),
$\text{n}=2\text{m}$
$=2\times4=8$
$\therefore \text{n}=8,\text{m}=4$

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Question 43 Marks

The angles of a triangle are in A.P. and the number of degrees in the least angle is to the number of degrees in the mean angle as 1 : 120. Find the angles in radians.

Answer

Let A, B & C be the angle of triangle ABC.
We are given that A, B & C are in A.P.
$\therefore$ Let A = a - d, B = a and C = a + d
According to the quation,
A + B + C = 180°
$\therefore$ a - d + a + a + d = 180°
⇒ 3a = 180°
⇒ a = 60°
Again,
$\frac{\text{least angle}}{\text{mean angle}}=\frac{1}{120^{\circ}}$
$\Rightarrow\ \frac{\text{a}-\text{d}}{\text{a}}=\frac{1}{120^{\circ}}$
$\Rightarrow\text{d}=\frac{119\text{a}}{120} $
$\Rightarrow \text{d}=\frac{119}{120}\times60^{\circ} $
$=\Big(\frac{119}{2}\Big)^{\circ}$
$=\Big(\frac{119}{2}\Big)\times\frac{\pi}{180}=\frac{119\pi}{360}\ \text{radians}$
Now,
$1^{\circ}=\frac{\pi}{180}\ \text{radians}$
$\text{B}=\text{a}-\text{d}=\frac{\pi}{3}\ \text{radians}$
$\text{A}=\text{a}-\text{d}=\frac{\pi}{3}-\frac{119\pi}{360}=\frac{\pi}{360}\ \text{radians}$
$\text{C}=\text{a}+\text{d}=\frac{\pi}{3}+\frac{119\pi}{360}=\frac{239\pi}{360}\ \text{radians}$

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Question 53 Marks

Find the length which at a distance of 5280m will subtend an angle of 1' at the eye.

Answer

Let AB be the rail road,
$\angle\text{AOB}=\theta=1'$
$\text{AB}=\text{AB}=\text{l}$
$\text{OA}=\text{OB}=\text{r}=5280\text{m}$
$1^{\circ}=\Big(\frac{1}{60}\Big)^{\circ}=\Big(\frac{1}{60}\times\frac{\pi}{180}\Big)^{\text{c}}$
Now,
we know that,
$\Rightarrow \Big(\frac{\pi}{180\times60}\Big)^{\text{c}}=\frac{\text{l}}{5280}$
$\Rightarrow \text{l}=\frac{5280\pi}{180\times60}=1.5365\text{m}$

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Question 63 Marks

Find the magnitude, in radians and degrees, of the interior angle of a regular.
Pentagon

Answer

General formula for in angles of polygon with n side $=\Big(\frac{2\text{n}-4}{\text{n}}\Big)\times90^{\circ}$
Pentagon has 5 sides,
Magnitude of the angle:
$=\frac{2\times5-4}{5}\times90^{\circ}$
$=\frac{6}{5}\times90^{\circ}$
$=180^{\circ}$
Now,
$\therefore 1^{\text{c}}=\frac{180}{\pi}$
And each angle of pentagon
$=\frac{2\times5-4}{5}\times\frac{\pi}{2}$
$=\Big(\frac{3\pi}{5}\Big)^{\text{c}}$
$\therefore 108,\Big(\frac{3\pi}{5}\Big)^{\text{c}}$

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Question 73 Marks

Find the magnitude, in radians and degrees, of the interior angle of a regular.
Duodecagon.

Answer

General formula for in angles of polygon with n side $=\Big(\frac{2\text{n}-4}{\text{n}}\Big)\times90^{\circ}$
Pentagon has 5 sides,
$\text{n}=12$
$\therefore$ Each angle $=\frac{2\times12-4}{12}\times90^{\circ}$
$=\frac{20}{12}\times90^{\circ}$
$=150^{\circ}$
Again,
Each angle $=\frac{2\times12-4}{12}\times\frac{\pi}{2}$
$=\frac{20}{12}\times\frac{\pi}{2}$
$=\Big(\frac{5\pi}{6}\Big)^{\text{c}}$
$\therefore 150^{\circ},\Big(\frac{5\pi}{6}\Big)^{\text{c}}$

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Question 83 Marks

One angle of a triangle $\frac{2}{3}\text{x}$ grades and another is $\frac{3}{2}\text{x}$ degrees while the third is $\frac{\pi\text{x}}{75}$ radians. Express all the angles in degrees.

Answer

Let $\theta_{1}$ $\theta_{2}$and $\theta_{3}$ be angles of a right angles triangle.
$\theta_{1}=\frac{2}{3}\times\text{gradiants}$
$\theta_{2}=\frac{3}{2}\times\text{degrees}$
$\theta_{2}=\frac{\pi\text{x}}{75}\times\text{radians}$
Now,
We have to express all the angles in degrees,
$\theta_{1}=\Big(\frac{3}{2}\text{x}\times\frac{90}{100}\Big)^{\circ}$
$=\frac{3}{5}\text{x}$
$\theta_{2}=\frac{\pi\text{x}}{75}\times\frac{180}{\pi}$
$=\frac{12\text{x}}{5}$
By angles property,
$\theta_{1}+\theta_{2}+\theta_{3}=180^{\circ}$
$\frac{3}{5}\text{x}^{\circ}+\frac{3}{2}\text{x}^{\circ}+\frac{12\text{x}}{5}=180^{\circ}$
$\Rightarrow\frac{9}{2}\text{x}^{\circ}=180^{\circ}$
$\Rightarrow \text{x}=40^{\circ}$
$\therefore\ \theta_{1}=24^{\circ},\theta_{2}=60^{\circ},\theta_{3}=96^{\circ}$

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Question 93 Marks

A rail road curve is to be laid out on a circle. What radius should be used if the track is to change direction by 25° in a distance of 40 metres?

Answer

Let AB be the rail road,
$\angle\text{AOB}=25^{\circ}$
$=25\times\frac{\pi}{108}=\Big(\frac{5\pi}{36}\Big)^{\text{c}}$
We konw that,
$\Rightarrow \angle\text{AOB}=\frac{\text{AB}}{\text{OA}}$
$\Rightarrow \frac{5\pi}{36}=\frac{40}{\text{r}}$
$\Rightarrow \text{r}=\frac{40\times36}{5\pi}$
$\Rightarrow \text{r}=\frac{288}{\pi}\ \text{meter}$ $\bigg[\therefore \pi = \frac{22}{7}\bigg]$
$\Rightarrow \text{r}=91.64\ \text{meter}$

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