Trigonometry for Angular Distance (Class- 9, Experience- 7)

• Topics
• MCQ Test
• Exercise Solution

[Exercise- page 178]

E.7.1 How many seconds equal 5°?

Solution:

Find how many seconds in 5°-

We know,
1° = 3600"
Therefore, 5° = (5 ✕ 3600)" = 18000"

So, 5° is 18000 second.  (Answer)

E.7.2 Construct the angles 30°, 360°, 380°, -20° and -420° using geometric ruler and protractor.

Solution:

Angles should be drawn using geometric ruler and protractor.

30° Draw:

(i) First draw a straight line AB by placing the geometric ruler.

(ii) If the center of the cantilever is placed at point A and the right side ends along AB, the semicircular portion of the cantilever must be upwards.

 
(iii) To mark a point C with a pencil along the line 30 from the subscript, mark the point C anti-clockwise from the ray AB.

 
(iv) Draw lines through origin A and marked point C so that they make an angle of 30°. Draw ∠BAC = 30°.



360° Draw:

360° angle is a full circle, so take any point on the circumference of the circle, it will be 360° away from the full circle.



380° Draw:

Here, 380° - 360° = 20°
That means, 360° is a full circle, then 380° angle is a 20° new rotation added to the full circle. Our drawing of 20° angle will suffice because 360° angle lies along the straight line OA.

 
(i) First draw a straight line OA by placing the geometric ruler.

 
ii) If the center of the cantilever is placed at point O and the right side ends along OA, the semicircular portion of the cantilever must be upwards.

 
(iii) To mark a point P with a pencil along the line 20 from the subscript, mark the point P anti-clockwise from the ray OA.

 
(iv) Draw lines through origin 0 and marked point P so that they subtend 360° + 20° = 380°.



-20° Draw:

(i) First draw a straight line OA by placing the geometric ruler.

 
(ii) If the center of the cantilever is placed at point O and the right side ends along OA, the semicircular portion of the cantilever must be downwards.

 
(iii) To mark a point P with a pencil along the line 45 from the subscript, mark the point P clockwise from the ray OA.

 
(iv) Draw lines through origin 0 and marked point P so that they make an angle of 45°. Draw ∠AOP = 45°.



-420° Draw:

Here, 360° - 420° = - 60°
That means, 360° is a full circle, then the angle 420° is a 60° new rotation added to the bottom of the full circle. Drawing an angle of 60° downwards will be sufficient for us since the angle of 360° lies along the straight line OA.

 
(i) First draw a straight line OA by placing the geometric ruler.

 
ii) If the center of the cantilever is placed at point O and the right side ends along OA, the semicircular portion of the cantilever must be downwards.

 
(iii) To mark a point P with a pencil along the line 60 from the subscript, mark the point P anti-clockwise from the ray OA.

 
(iv) Draw lines through origin 0 and marked point P so that they subtend -360° - 60° = 420°.

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E.7.3 Construct the angles 60°, 90°, 180°, 200°, 280°, 750°, -45°, -400° at their standard position using geometric ruler and protractor. Determine whether these are quadrant angles or quadrantal angles. State which quadrant the angles are in.

Solution:

Draw the ideal position of the corners. Determine whether they are quadrants or quadrantal angles. In which quadrant the angles are to be specified.

Using a ruler and chisel the corners are drawn in the ideal position as shown in the figure below.

60°,

60° is a quadrant angle. The angle is in the first quadrant.



90°,

 
90° is a quadrantal angle. It lies on the positive y-axis. Hence it is not included in any quadrant. This is an angle known as quadrantal angle.



180°,

 
180° is a quadrant angle. The angle is in the second quadrant.

[180° is considered a quadrant angle because it falls within the second quadrant, which is defined as the region where the x-axis value is negative and the y-axis value is positive. Thus, 180° is located in the second quadrant based on its position relative to the coordinate axis.]

200°,

 
200° is a quadrant angle, the angle is in the third quadrant.

280°,

 
280° is a quadrant angle. The angle is in the fourth quadrant.



750°,

 
To find the equivalent angle between 0° and 360°, subtract 360° from 750°:
750° - 360° = 390°

So, an angle of 750° is subtended by 390°. However, since an angle of 390° is greater than 360°, it does not belong to any specific quadrant.



–45°,

 
-45° is considered a quadrant angle. It falls into the fourth quadrant.



–400°

To find the equivalent angle between 0° and 360°, add 360° to -400°:
-400° + 360° = -40°

So, -400° equals -40°. However, since -40° is negative and falls in the fourth quadrant, we can say that -400° is in the fourth quadrant.

E.7.4 Find the values: cos135°, cot120°, tan390°, sin(–30°), sec300°, csc(–570°)

Solution:

Values to be found: cos135°, cot120°, tan390°, sin(–30°), sec300°, csc(–570°)


    cos135°
= cos(180° - 45°)
= -cos45°
= -   1√ 2    [Since, cos45° =   1√ 2 ]

[135° is in the second quadrant. Its reference angle in the first quadrant is 45°. Cosine of 45° is =   1√ 2 ]



    cot120°
= cot(180° - 60°)
= -cot60°
= -    1√ 3    [Since, cot60° =   1√ 3  ]



    tan390°
= tan(360° + 30°)
= tan30°
=   1√ 3    [Since, tan30° =   1√ 3 ]



    sin(–30°)
= -sin30°
= -  1  2   [Since, sin30° =  1  2]



    sec300°
= sec(360° - 60°)
= sec60°
= 2     [Since,sec60° = 2]



    csc(–570°)
= csc570°
= csc(540° + 30°)
= csc30°
= 2  [Since, csc30° = 2]

E.7.5 Find the trigonometric ratios of the angles at standard position where the points A(2, 3), B(–3, 1), C(–4, –4), D(1, –2), and E(–2, 0) are on the terminal rays.

Solution:

Find the trigonometric ratio of angles subtended by points A(2, 3), B(–3, 1), C(–4, –4), D(1, –2), E(–2,0) at standard positions. To do-


A(2, 3)

Given,
Adjacent side, x = 2,
Opposite side, y = 3 and
So, Hypotenuse, r = √ 2² + 3²  = √ 13 

Here, he trigonometric ratios are-

sinθ =  y  r =    3√ 13 

cosθ =  x  r =    2√ 13 

tanθ =  y  x =  3  2

cotθ =  x  y =  2  3

secθ =  r  x = √ 13   2

cscθ =  r  y = √ 13   3



B(–3, 1)

Here,
Adjacent side, x = -3,
Opposite side, y = 1 and
So, Hypotenuse, r = √ (-3)² + 1²  = √ 10 

Here, he trigonometric ratios are-

 
sinθ =   y  r =   1√ 10 

cosθ =  x  r  = -   3√ 10 

tanθ =  y  x = -  1   3

cotθ =  x  y = -3  1 = -3

secθ =  r   x = √ 10   -3

cscθ =  r   y = √ 10    1 = √ 10



C(–4, –4)

Here,
Adjacent side, x = -4,
Opposite side, y = 4 and
So, Hypotenuse, r = √ (-4)² + 4²)  = √ 32  = √ 16 ✕ 2   = 4√ 2 

Here, he trigonometric ratios are-

 
sinθ =  y  r =    4 4√ 2  =   1√ 2 

cosθ =  x  r = -    4 4√ 2  = -   1√ 2 

tanθ =  y  x =   4   -4  = -1

cotθ =  x  y = -   4    4 = -1

secθ =  r  x =  4√ 2   -4 = -√ 2 

cscθ =  r  y =  4√ 2    4 = √ 2 

D(1, –2)

Here,
Adjacent side, x = 1,
Opposite side, y = -2 and
So, Hypotenuse, r = √ 1² + (-2)²  = √ 5 

Here, he trigonometric ratios are-

sinθ =  y  r = -   2√ 5 

cosθ =  x  r = 1√ 5 

tanθ =  y  x =  -2   1 = -2 

cotθ =  x  y =  1 -2 = -  1  2

secθ =  r   x = √ 5   1 = √ 5 

cscθ =  r   y = √ 5   2



E(–2, 0)

 
Here,
Adjacent side, x = -2,
Opposite side, y = 0 and
So, Hypotenuse, r = √ (-2)² + (-1)²  = 2

Here, he trigonometric ratios are-

sinθ =  y  r =  0  2 = 0

cosθ =  x  r = -  2  2 = -1

tanθ =  y  x =  0 -2 = 0

cotθ =  x  y = -  2  0 = undefined

secθ =  r   x =  2 -2 = -1

cscθ =  r   y =  2  0 = undefined

E.7.6 Express the following points using r and tanθ. 

a. A(3, –2)

b. B(–2, –1)

c. C(–4, 0)

Solution:

The following points are to be expressed in terms of r and tanθ:

a. A(3, –2)

Here,
Adjacent side, x = 3,
Opposite side, y = -2
So, Hypotenuse, r = √ 3² + (-2)²  = √ 13 
এবং,
tanθ =  y  x = -2 3

Therefore, (r, θ) = (√ 13 , -  2  3) [Answer]



b. B(–2, –1)

Here, x = -2, y = -1
So, Hypotenuse, r = √ (-2)² + (-1)²  = √ 5 
এবং,
tanθ =  y  x =   -1  -2  =  1  2 

Therefore, (r, θ) = (√ 5 ,  1  2) [Answer]



c. C(–4, 0)

Here, x = -4, y = 0
So, Hypotenuse, r = √ (-4)² + 0²  = 4
এবং,
tanθ = yx =   0 -4  = 0

Therefore, (r, θ) = (4, 0) [Answer]

E.7.7 Express in radian: 

a.75°30’ 

b. 45°44’43’’

c. 60°30’15’’

Solution:

Express in radians:

a.75°30'

    75°30'

= 75° + ( 30  60)°  [We know, 1° = 60’]

= 75° + ( 1  2)°

= (75 ✕ 2 + 1        2)°

= ( 151    2)°

= ( 151    2) ✕    π    180 radians  [We know, 1° =    π   180 radians]

= 151π  360  radians



b. 45°44'43"

    45°44'43"

= 45° + ( 44  60)° + (   43    3600)°   [We know, 1° = 60’ and 1° = 3600”]

=    π   180 (45 +  44  60 +    43    3600) radians    [We know, 1° =   π180 radians]

=    π    180 ✕  (45 ✕ 3600 + 44 ✕ 60 + 43)                 3600  radians

=    π    180 ✕  (162000 + 2640 + 43)             3600  radians

=    π    180 ✕  164683    3600  radians

=  164683π   648000  radians




c. 60°30'15"

    60°30'15"

= 60° + ( 30  60)° + (   15   3600)°   [We know, 1° = 60’ and 1° = 3600”]

= 60° + ( 1  2)° + (   1    240)°

=   π180 (60 +  1  2 +    1    240) radians   [We know, 1° =   π180 radians]

=   π180 ✕ (60 ✕ 240 + 1 ✕ 120 + 1)                240  radians

= π(14400 + 120 + 1)       240 ✕ 180  radians

=  14521π   43200  radians

E.7.8 Express in degrees: 

a. 4π25 radian 

b. 1.3177 radian 

c. 0.9759 radian

Solution:

Express in degrees:

 
a. 4π25 radians

 4π  25 radians

= ( 4π  25 ✕  180    π )°  [We know, 1 radians = 180°/π]

= ( 4π  25 ✕  180    π )°

=  4 ✕ 36    5

= 28.8°




b. 1.3177 radians

    1.3177 radians

= (1.3177 ✕   180    π)°  [We know, 1 radians =   180    π]

= (1.3177 ✕     180   3.1416)°  [We know, π = 3.1416]

= 75.4984° (approx.)




c. 0.9759 radians

    0.9759 radians

= (0.9759 ✕  180    π )°  [We know, 1 radians =   180    π]

= (0.9759 ✕    180 3.1416)°  [We know, π = 3.1416]

= 55.9148° (approx.)

E.7.9 Radius of the Earth is 6440 km. If the positions of Teknaf and Tetulia make an angle of 10°6’3’’ at the center of the Earth, what is the distance from Teknaf to Tetulia?

Solution:

Find the distance from Teknaf to Tatulia-

Here,
The radius of the earth, r = 6440 km.
The angle subtended at the center of the Earth by the positions of Teknaf and Tatulia, θ = 10°6'3"

Now,

θ = 10°6'3"

   = 10° + (  6   60)° + (    3 3600 )°

   = 10° + (  1   10)° + (11200)°

   = (1200 ✕ 10 + 120 + 1           1200)°

   = ( 12121   1200)°

   =   π180 ✕  12121  1200 radians   [We know, 1° =   π180 radians]

   =  12121π  216000 radians

Now, the distance between Teknaf and Tatulia, 

S = rθ
   = 6440 ✕ 12121π216000
   = 1135.328 km (approx.)

So, the distance between Teknaf and Tatulia 1135.328 km (approx.)। (Answer)  

E.7.10 Radius of the Earth is 6440 km. Suppose, two satellites are positioned above the earth in such a way that they make an angle of 33” with the center of the Earth. What is the distance between the two satellites?

Solution:

To find the distance between the two satellites-

Given,
he radius of the earth, r = 6440 km
and the angle between the two satellites at the center of the earth, θ = 33"

Now,

θ = 33"
   = (  333600)°
   =   π180 ✕    33 3600 radians [We know, 1° =   π180 radians]

So, the distance between the two satellites,

S = rθ

   = 6440 ✕   π180 ✕   333600 km

   = 1 km (approx.)