Measurement of Angles

1. Two angles of a triangle are 60g and 130g. Find the remaining angle in degree.
Solution: Here, 
Let the remaining angle in grade be x. 
We have, 
      60^g + 120^g + x^g = 200^g [∴Sum of angles of triangle in grade is 200ᵍ]
i.e.   x = (200 – 180)^g
∴    x = 20^g.
Now,
        x = 20 × 9^o10
           = 18°.
Thus, the remaining angle in degree is 18°.

2. The three angles of a triangle are  (2x3 )^g , (3x2 )^ο and (2x3 )^c. Find the measure of each angle in degrees.
Solution: Given angles are:
  ∡ A = (2x3)^g = (2x3 × 910)^ο = (9x15)^ο = (3x5)^ο
  ∡ B = (3x2)^ο 
  ∡ C = (πx75)^c = (πx75 ×180π)^ο = (12x5)^ο

Here, ∡ A, ∡ B and ∡ C are angles of a triangle.
∴ 3x5 +  3x2 +  12x5 = 180°
⇒ 9x2 = 180°
⇒ x = 180° ×29 = 24°.

∴      ∡B = 3x2 = 3 × 40°2 = 60°
And ∡C = 12x5 = 12 × 40°5 = 96°.

3. The sum of number of degrees and grades of an angle is 152°. Find the angle in degrees.
Solution: 
Let, the required angle in degree = x`
Now, x^o = 109x grade.
By the question,
     x+109x=152^o
⇒ 9x+10x9=152^o
⇒ 19x=1368^o
 ⇒      x=136819
         =72^o.
Hence, the required angle in degree is 72^o.

4. In the given figure, O is the centre of a circle. If AOB = 3π5c OA = 10 cm, find the length of arc ACB.

Solution: 
Here,
       Radius of the circle = r = 10 cm
       Central angle subtended by arc ACB = ∡ACB
                                                                   = 2πc-3π5c
                                                                                  =10π-3π5c
                                                                                  =7π5^c
Now,
         Length of the arc ACB = ∡ ACB⨉ r
                                              = 7π5×10cm
                                             =14πcm
                                             =14×227cm
                                              =44 cm.
Hence, the required length of arc ACB is 44 cm.

5. ACB is an arc of the circle with centre O. If AO = OB = 5.6 cm and  AOB = 3π4c. Determine the length of the arc ACB.

Solution: 
Here,
        Radius of the circle = r = 5.6 cm
        Central angle subtended by arc ACB = ∡AOB
                                                                    = 2πc-3π4^c
                                                                                    =8π-3π4^c
                                                                                    =5π4^c

Now,
       Length of the arc ACB = ∡AOB ⨉ r
                                             = 5π4×5.6cm
                                            =7πcm
                                            =7×227
                                             =22 cm.

Hence, the required length of arc ACB is 22 cm.

6.  The radius of a circle is 7 cm. Find the angle in degrees subtended by an arc of 323 cm at the centre of the circle.
Solution: 
Here,
     Length of the arc = l = 323cm=113 cm
     The radius of the circle = r = 7 cm
   Now, Angle subtended by an arc = lr
                                                     =11^c37
                                                     =1121^c
                                                                              =1121×180^oπ
                                                     =30°.

Therefore, angle subtended by arc = 30^o.

7. The radius of a circle is 5.6 cm. Find the angle in degrees subtended by an arc of 22 cm at the centre of the circle.
Solution: 
Here,
        Length of the arc = l = 22 cm
        The radius of the circle = r = 5.6 cm
      Now,
      Angle subtended by an arc = lr
                                                 =225.6^c
                                                                        =1121×180π
                                                 =225°.
Hence, the required angle subtended by an arc is 225°.

8. The radius of a circle is 21 cm. An arc of the circle subtends an angle of 60' at the centre. Find the length of the arc.

Solution: 

Here,
       Radius of the circle = 21 cm
       Central angle subtended by arc = 60'
= 60×π180^c=π3^c
Now,
     Length of the arc = Central angle ⨉ Radius
                                  = π3^c×21cm
                                 =7πcm
                                 =7×227cm
                                  =22 cm.
Hence, the required length of the arc is 22 cm.

9. The given figure is a part of the circle with centre O and the arc AB. If OA = 14 cm and AOB =3π4c, calculate the length of the arc AB. Also, find the perimeter of the figure OAB.

Solution:
Here, the radius of the section of the circle = r = OA = 14 cm
Central angle subtended by arc AB = AOB = 3π4c
Now, length of the arc AB = AOB*r =
 3π4c×14=10.5π=10.5×227=33 cm
Again, the perimeter of the section = Arc AB + OA + OB = 33 + 14 + 14 = 61 cm

10. Find the value of arc ACB in the given figure alongside.

Solution:
Here, the radius of the circle = r = OA = 10 cm
Central angle subtended by arc ACB = AOB = 2πc-2π5c=10π-2π5c=8π5c
Now, length of the arc ACB = AOB*r = 8π5×10=16π=7×227=50.28 cm
Hence, the required length of arc ACB is 50.28 cm.

11. Find the value of θ in the given figure alongside.

Solution: Here,
     Length of the arc AB = 44 cm
     Radius of the circle = r = 21 cm
     Angle = ?
Now,
      θ=Arc ABr
        =4421
       =2.1c .
Hence, the required value of θ=2.1c.

12. One angle of a triangle is 27°. If the ratio of remaining two angles is 8:9, find all angles of the triangle in grades.
Solution: Let the angles are 8x+9x. and one angle 27°


we know that ,
      9x+8x+27 = 180°
⇒ 17x = 180-27
⇒ x = 153°17 = 9°
∴ ∢ A = 9x
           = 9 × 9°
           = 81°
           = (81×109)^g
           = 90^g
∢ B = 8x
       = 8×9°
       = 72°
       = (72×109)^g
       = 80^g
and ∢ C = 27°
              = (27×10/9) °
              = 30^g.

13. Two angles of a triangle are in the rato of 3 :8 and the third is 9. Find all angles in degrees.
Solution: Given,
3rd angle is 9π c20 = 9π c20 ×180π c = 81°
Let 1st and 2nd angle are 3x & 8x.
Then,
     3x + 8x +81 = 180°
⇒ 11x = 99°
⇒ x = 9°.
∴  1st angle = 3x = 3×9° = 27°
    2nd angle = x = 8 ×9° = 72°
& 3rd angle = 81°.

14. One angle of a triangle is 45°.If the ratio of remaining two angles is 8:7, find all angles of the triangle in grades.
Solution: One angle of a triangle is 45°.
Let the other angle are 8x and 7x.
Then,
     8x+7x+45° = 180°
⇒ 15x = 135°
⇒ x= 9 °
∴1st angle = 8x = 8 ×9° = (109× 8×9)^g = 80^g
2nd angle = 7x = 7×9° = (109× 7×9)^g = 70^g
and last angle = 45° = (109× 45)° = 50^g.

15. The three angles of a triangle are (9x/2)° , 6x^g    and (9πx200) c. Find the measure of each angle in grades.
Solution: Given angles are, 9x°2 = (9x2×109) ^g = 5x^g
  6x^g   and (9π x200)^c = ( 9π x200 ×200π)^g = 9x^g
We have
      5x+6x+9 = 200^g (in grade)
⇒ 20x = 200^g
⇒ x = 10^g.
∴ 1st angle = 5x^g = 5× 10^g = 50^g
   2nd angle = 6x^g = 6× 10^g = 60^g and
   3rd angle = 9x = 9× 10^g = 90^g.

16. The three angles of a triangle are (2x3)g , (3x2) ° and (πx75)^c  . Find the measure of each angle in grades.

Solution: Given angles are 2x g3 
 3x °2 = (3x2×109) ^g
        = 5x g3 and
 (πx75)^c= (πx75×π200 ) (200π)^g
         = 8x3 ^g
We have,
 2x g3 + 5x g3 + 8x g3 = 200^g (in grade)
⇒ 15x = 200×3
⇒ x = 200×315
        = 40^g.
∴ 1st angle = 2x3 = 2 ×403 = 80 g3
2nd angle = 5x3 = 5 ×403 =  200 g3 and
Last angle = 8x3 = 8 ×403 = 320 g3 .

17. One angle of a triangle is two third of a right angle and the greater of the other two exceeds the less by 20^g .Find all the angles in degrees.
Solution: Here,
One angle is equal to 23× 90° = 60°
If 2nd angle is x & then 3rd angle is
   x + 20 ^g = x + 20 × 910° = x + 18°.
We have,
    60 + x + x + 18° = 180°
⇒ x = 51°
∴ Angles are 60°, 51° & (51^o+18^o) = 69°.

18. Three angles of a triangles are (a-d) ° , a ° and (a+d) ° , the number of degree of the greatest angle is 80. Find all the angles in radian.
Solution: Let the given angles are,
      a - d, a & a + d, where a + d is the greatest angle.
i.e  a + d = 80...............(i)
We have,
     (a - d) + a + (a + d) = 180°
⇒ 3a = 180°
⇒ a = 60 °
Put, a = 60° in (i), we get,
    d = 80° - 60° = 20°.
∴ The angles are a - d = 60° - 20°
                                    = 40°
                                    = (40×π180)^c
                                    = 2 π c9
a° = 60°
    = (60 ×π180)c
    = π c3
and, a + d = 80 °
                 = (80 ×π180)^c
                 = 4π c9.

19. If an angle of a triangle exceeds the second by 20° and the circular measure of the third angle exceeds that of the second by  ^c,find the angles in degree.
Solution: Let the 2nd angle be x ^c, then
1st angle = x ^c +20° = x + 20° X π c180° = x + π c9 and,
3rd angle = x + π c18°
We know,
      x ^c + (x^c + π9 )+ x + (π c18) = π (in radian π^c = 180°)
⇒ 3x = π - π9 - π18
⇒ 3x = 5π6
⇒ x = 5π18
∴ Required angles are,
     x +18° = x + π9 = 5π9 + π18 = 11π c18
    x^c= 5π c18 and
     x + π18 = 5π18 + π18= 6π c18 = π c3.

20. In a right angled triangle, the number of degrees in one acute angle is equal to the number of grades in the other acute angle. Find the acute angles in radians.
Solution: Let x be the angle in degree then another angle is x in grade,
∴ x grade = (x ×910)°
                = 9x10
In right angled triangle,
x + 9 x10 = 90°
⇒ 19 x10 = 90°
⇒ x = 900 °19
∴ 1st acute angle =  900 °19× π c180 = 5π c19 
   2nd acute angle = ( 9102× 90019 × π180 )^c = 9π c38.

21. The number of degrees in an angle of a triangle is to the number of grades in the second angle is to the number of radians in the third as 72:70 : π4. Find all the angles in degrees.
Solution: Let the common ratio be x, then
First angle = 72x°
Second angle = 70x^g = (70x× 910)° = 63x°
Third angle = π4x = π4x ×180°π = 45 x°.
We know,
72x + 63x + 45x = 180°
⇒ 180x = 180°
⇒ x = 1°.
∴ First angle = 72x = 72 × 1° = 72°
   Second angle = 63x = 63 × 1° = 63° and
   Third angle = 45x = 45 × 1° = 45°.