(Each question 3 marks)

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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MATHS (Each question 3 marks)

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