Soln:
Here, 27o = 27× 60' = 1620'

27o38' = (1620 + 38)' = 1658'

Then, 1658' = (1658 × 60)" = 99480"

∴ 27o38'42" = (99480 + 42)" = 99522". Ans.

Soln:

Here, 60o = 60× 60' = 600'

3600' = (3600× 60") = 216000"

∴ 60o = 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 = 0.17o

∴ 20o 10' 12" = (20 + 0.17)o = 20.17o . 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 = 0.1694o

∴ 20o 10' 10" = (20 + 0.1694)o = 20.1694o. 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 = 0.2286o

∴ 57o 13' 43" = ( 57 + 0.2286)o = 57.2286o 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 = 0.2583o

∴ 60o 15' 30" = ( 60 + 0.2583)o = 60.2583o 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 = 0.1425g

∴ 6g 14' 25" = 6 + 0.1425g = 6.1425g 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 = 0.0285g

∴ 80g 2' 85" = ( 80 + 0.0285)g = 80.0285g. 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 = 0.059g

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

Soln:

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

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

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

Soln:

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

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

= 22.73607o = 22o 44' 9.85". Ans.

Soln:

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

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

= 27.262777o = 27o 15' 46" Ans.

Soln:

Here,26o46' = (26 +$$\frac{46}{60}$$)o = ( 26 + 0.766)o = 26.7666o

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

= 29g 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 = 27o

If the third angle = xo

We know sum of angles of the triangle = 180o

or, xo + 27o + 90o = 180o

or, xo = 180o - 117o = 63o

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

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

∴ Third angle of the triangle = 70g. Ans.

Soln:

Here given,

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

Ratio of the remaining two angles is 1: 3

Let, these two angles be kg and 3kg..

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

80 + k + 3k = 200

or, 4k = 200 - 80

or, 4k = 120

∴ k = $$\frac{120}{4}$$ = 30

∴ First angle = 80g

Second angle = k = 30g

and Third angle = 3k = 3× 30 = 90g

∴ Three angles = 30g 80g90g. 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.

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 = 180o.

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, $$\frac{1^c}{180^o}$$ =$$\frac{r}{πr}$$

∴ 1c = $$\frac{180^o}{Π}$$

i.e.πc = 180o

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