Soln:

Here, 27^{o} = 27× 60' = 1620'

27^{o}38' = (1620 + 38)' = 1658'

Then, 1658' = (1658 × 60)" = 99480"

∴ 27^{o}38'42" = (99480 + 42)" = 99522". Ans.

Soln:

Here, 60^{o} = 60× 60' = 600'

3600' = (3600× 60") = 216000"

∴ 60^{o} = 216000" Ans.

Soln:

Here, 10" =(\(\frac{12}{60}\))

10' 12" =(10 +\(\frac{12}{60}\))' = (10 + 0.2)' = 10.2'

10.2' =(\(\frac{10.2}{60}\))^{o} = (\()\frac{102}{600}\)^{o} = 0.17^{o}

∴ 20^{o} 10' 12" = (20 + 0.17)^{o} = 20.17^{o} . Ans.

Soln:

Here, 10" = (\(\frac{10}{60}\))' = (\(\frac{1}{6}\))'

10' 10" = (10 + \(\frac{1}{60}\)) =(\(\frac{60 + 1}{6}\))' =\(\frac{61'}{6}\)

\(\frac{61'}{6}\) =(\(\frac{61}{6}\)× \(\frac{1}{60}\))^{o} =(\(\frac{61}{360}\))^{o} = 0.1694^{o}

∴ 20^{o} 10' 10" = (20 + 0.1694)^{o} = 20.1694^{o}. Ans.

Soln:

Here, 43" =(\(\frac{43}{6}\))'

13'43" =(13 + \(\frac{43}{60}\))' =(\(\frac{823}{60}\))'

(\(\frac{823}{60}\))' =(\(\frac{823}{60}\)×\(\frac{1}{60}\))^{o} = 0.2286^{o}

∴ 57^{o} 13' 43" = ( 57 + 0.2286)^{o} = 57.2286^{o} Ans.

Soln:

Here, 30" =(\(\frac{30}{60}\))' =(\(\frac{1}{2}\))'

15'30" = (15 + \(\frac{1}{2}\))' =(\(\frac{31}{2}\))'

(\(\frac{31}{2}\))' =(\(\frac{31}{2}\)×\(\frac{1}{60}\))^{o} = 0.2583^{o}

∴ 60^{o} 15' 30" = ( 60 + 0.2583)^{o} = 60.2583^{o} Ans.

Soln:

Here, 25" = (\(\frac{25}{100}\))' =(\(\frac{1}{4}\))'

14' 25" = ( 14 + \(\frac{1}{4}\))' =(\(\frac{57}{4}\))'

(\(\frac{57}{4}\))' =(\(\frac{57}{4}\) × \(\frac{1}{100}\))^{g} =(\(\frac{57}{400}\))^{g} = 0.1425^{g}

∴ 6^{g} 14' 25" = 6 + 0.1425^{g} = 6.1425^{g} Ans.

Soln:

Here, 85" = (\(\frac{85}{100}\))' = (\(\frac{17}{20}\))'

2'85" = ( 2 + \(\frac{17}{20}\))' = (\(\frac{57}{20}\))'

(\(\frac{57}{20}\))' =(\(\frac{57}{20}\)×\(\frac{1}{100}\))^{g} =(\(\frac{57}{2000}\))^{g} = 0.0285^{g}

∴ 80^{g} 2' 85" = ( 80 + 0.0285)^{g} = 80.0285^{g}. Ans.

Soln:

Here, 90" = (\(\frac{90}{100}\))' = (\(\frac{9}{10}\))'

5' 90" = ( 5 + \(\frac{9}{10}\))' = (\(\frac{59}{10}\))'

(\(\frac{59}{10}\))' =(\(\frac{59}{10}\)×\(\frac{1}{100}\))^{g} = 0.059^{g}

∴75^{g} 5' 90" = (75 + 0.059)^{g} = 75.059^{g}. Ans.

Soln:

Here, 70" = (\(\frac{70}{100}\))' =(\(\frac{7}{10}\))'

(\(\frac{7}{10}\))' =(\(\frac{7}{10}\)×\(\frac{1}{100}\))^{g} = (\(\frac{7}{1000}\))^{g} = 0.007^{g}

∴ 65^{g} 70" = ( 65 + 0.007)^{g} = 65.007^{g} Ans.

Soln:

Here, 25^{g} 26' 23" = 25.2623^{g} [ 25 +\(\frac{26}{100}\) + \(\frac{23}{10000}\) = 25+0.26+0.0023]

Changing into grade

25.2623^{g} =(25.2623× \(\frac{9}{10}\))^{o}

= 22.73607^{o} = 22^{o} 44' 9.85". Ans.

Soln:

Here,30^{g} 29' 19.75" = 30.291975^{g} [ 30 + \(\frac{29}{100}\) + \(\frac{19.75}{10000}\) = 30 + 0.29 + 0.001975]

30.29195^{g} = (30.291975× \(\frac{9}{10}\))^{o}

= 27.262777^{o} = 27^{o} 15' 46" Ans.

Soln:

Here,26^{o}46' = (26 +\(\frac{46}{60}\))^{o} = ( 26 + 0.766)^{o} = 26.7666^{o}

26.7666^{o} = ( 26.7666×\(\frac{10}{9}\))^{g} = 29.74066^{g}

= 29^{g} 74' 6.6" Ans.

Solution:

Let the no. of sides of the regular polygon be n.

Then, exterior angle = \(\frac{}{}\) × interior angle.

i.e. \(\frac{360}{n}\) = \(\frac{1}{4}\) × \(\frac{(n - 2) × 180°)}{n}\)

or, \(\frac{360}{n}\) = \(\frac{45(n - 2)}{n}\)

or, n - 2 = \(\frac{360}{45}\)

or, n - 2 = 8

\(\therefore\) n = 10

Solution

When the clock strikes half past three.

The hour hand at 3 and minute hand at 6.

So, for real calculation,

The hour hand moves 30 minutes away from 3.

So, In 1 hour, the hour hand makes 30°.

\(\therefore\) In \(\frac{}{}\) hour, the hour hand makes 30 × \(\frac{1°}{2}\) = 15°

Also, the angle between the hour and minute hand when the hand is at 3 and 6 is 90°.

So, the angle between the two hands at half past three 90° - 15° = 75°.

Soln:

Here, given

One angle of a right angle triangle = \(\frac{3}{10}\)× 90 = 27^{o}

If the third angle = x^{o}

We know sum of angles of the triangle = 180^{o}

or, x^{o} + 27^{o} + 90^{o} = 180^{o}

or, x^{o} = 180^{o} - 117^{o} = 63^{o}

but, 1^{o} = (\(\frac{10}{9}\))^{g} so,

63^{o} = ( 63 ×\(\frac{10}{9}\))^{g} = 70^{grd}

∴ Third angle of the triangle = 70^{g}. Ans.

Soln:

Here given,

One angle of triangle = 72^{o} = ( 72× \(\frac{10}{9}\))^{g} = 80^{g}

Ratio of the remaining two angles is 1: 3

Let, these two angles be k^{g} and 3k^{g.}.

We know, the sum of three angle of triangle = 200^{g}

80 + k + 3k = 200

or, 4k = 200 - 80

or, 4k = 120

∴ k = \(\frac{120}{4}\) = 30

∴ First angle = 80^{g}

Second angle = k = 30^{g}

and Third angle = 3k = 3× 30 = 90^{g}

∴ Three angles = 30^{g} 80^{g}90^{g}. Ans.

**Proof:**

Let O be the centre and OA = r be the radius of a circle. Let AP be an arc which subtends an angle AOP on the centre O of the circle and the length of which is equal to the radius i.e. AP = r.

Then, by definition of radian,∠AOP = 1 radian (1^{c}).

Now, producing AO up to B so that the arc length APB =\(\frac{1}{2}\) of the circumference =\(\frac{1}{2}\)× 2Πr =πr

Where AOB = diameter = 2r.

and∠AOB = 180^{o}.

Now, by the theories of geometry,

the angles at the centre of the circle are proportionalto the arcs on which they stand:

So, \(\frac{∠AOP}{∠AOB}\) =\(\frac{arc AP}{arc APB}\)

or, \(\frac{1^c}{180^o}\) =\(\frac{r}{πr}\)

∴ 1^{c} = \(\frac{180^o}{Π}\)

i.e.π^{c} = 180^{o}

Since the value of 1^{c} =\(\frac{180^o}{Π}\) is independent of r, so the radian is a constant angl. Proved