Trigonometry in Measurement (Class- 9, Experience- 6)

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• MCQ Test
• Exercise Solution

[Exercise- page 156]

E.6.1 If cosθ = 34 , find out the other ratios of the angle θ.

Solution:

If cosθ =  3  4, need to find out the other ratios of the angle θ.

Given,
cosθ =  3  4
sinθ = ?
tanθ = ?
cotθ = ?
secθ = ?
cscθ= ?


We know,

sin²θ + cos²θ = 1

or, sin²θ = 1 - cos²θ

or, sin²θ = 1 - ( 3  4)² [put down the value]

or, sin²θ = 1 -  916

or, sin²θ =  16 - 9     16

or, sin²θ =  716

∴ sinθ = √ 7   4


Again,

tanθ = sinθcosθ

or, tanθ = sinθ ÷ cosθ

or, tanθ = √ 7   4 ÷  3  4

or, tanθ = √ 7   4 ✕  4  3

∴ tanθ =  √ 7     3


Again,

cotθ =   1tanθ
                     1
or, cotθ = --------
                  √ 7   3

∴ cotθ =   3√ 7 


Again,

secθ =   1cosθ

                    1
or, secθ = ------
                    3  4

∴ secθ =  4  3


Again,

cscθ =   1sinθ

                    1
or, cscθ = ------
                  √ 7   4
∴ cscθ =   4√ 7 


So, the other ratios of the angle θ are, sinθ = √ 7   4, tanθ = √ 7   3, cotθ =   3√ 7 , secθ =  4  3, cscθ =   4√ 7   (Answer)

E.6.2 If 12 cotθ = 7, what is the value of cosθ and cscθ?

Solution:

If 12cotθ = 7, need to find value of cosθ and cscθ.

Given,

12cotθ = 7

or, cotθ =   7   12

or,   1tanθ =   7  12

or, tanθ =  12   7

or, sinθ°cosθ° =  12   7 

or, 7sinθ = 12cosθ

or, (7sinθ)² = (12cosθ)²  [square both sides]

or, 49sin²θ = 144cos²θ....(i)

or, 49(1 - cos²θ) = 144cos²θ      [Since, sin²θ + cos²θ = 1]

or, 49 - 49cos²θ = 144cos²θ

or, 49 = 144cos²θ + 49cos²θ

or, 144cos²θ + 49cos²θ = 49

or, 193cos²θ = 49

or, cos²θ =   49  193

or, cos²θ =   √ 49   √ 193 

∴ cosθ =      7√ 193 


Again, from equation (i) no., we get,

49sin²θ = 144cos²θ

or, 49sin²θ = 144(1-sin²θ)  [Since, sin²θ + cos²θ = 1]

or, 49sin²θ = 144 – 144sin²θ

or, 49sin²θ + 144sin²θ = 144

or, 193sin²θ = 144

or, sin²θ =  144  193

or, sin²θ =  144  193

or,    1sin²θ =  193  144    [Since, sin²θ =     1 csc²θ ]

or, csc²θ =  193  144

or, cscθ =  √ 193  √ 144  

∴ cscθ =  √ 193      12

So, cosθ =  √ 193  √ 144  

∴ cscθ =  √ 193      12  and cscθ =     7√ 193     (Answer)

E.6.3  In a right angled triangle ΔABC, ∠B = 90°, AC = 12 cm, BC = 13 cm and ∠BAC = θ. Find out the value of sinθ, secθ and tanθ.

Solution:

In a right angled triangle ΔABC, ∠B = 90°, AC = 12 cm, BC = 13 cm and ∠BAC = θ. Find out the value of sinθ, secθ and tanθ. [The length of the hypotenuse of a triangle cannot be less than the other side, so side AC is resolved by taking the hypotenuse]

If ∠B = 90°, AC = 13 cm, BC = 12 cm and ∠BAC = θ of right triangle ∆ABC , find the values of sinθ, secθ and tanθ.

According to the question, the figure drawn is,

In a right triangle ∆ABC, ∠ABC = 90°,
AC = 12 cm,
BC = 13 cm and
∠BAC = θ,

sinθ = ?
secθ = ?
tanθ = ?

According to the Pythagorean theorem-

AC² = AB² + BC²
or, AB² = AC² - BC²
or, AB² = 132 - 122
or, AB² = 169 - 144
or, AB² = 25
or, AB² = 25
or, AB² = √ 25 
∴ AB = 5


Here,
sinθ =    opposite  hypotenuse

or, sinθ =   AB    AC

∴ sinθ =   5   13


Again,

secθ =   hypotenuse    adjacent 

or, secθ =   AC    BC

∴ secθ =   13    12


and
tanθ =   opposite    adjacent

or, tanθ =   AB    AC

∴ tanθ =   5   12


So, sinθ =   5   13, secθ =  13  12, and tanθ =   5   12   (Answer)

E.6.4  If θ = 30°, prove that 

            i) cos2θ = 1 - tan²θ1 + tan²θ

 

            ii) tan2θ =   2tanθ1 - tan²θ

Solution:

If θ = 30°, show that,

(i) cos 2θ =  1 - tan²θ   1 + tan²θ


Given,
θ = 30°

 
Need to be shown that,

cos 2θ =  1 - tan²θ   1 + tan²θ

Here, Right side-

=  1 - tan²θ   1 + tan²θ

=  1 - tan²30°  1 + tan²30°

= (1 - tan²30°) ÷ (1 + tan²30°)

= {1 – (  1√ 3 )²} ÷ {1 + (  1√ 3 )²}  [Since, tan30° =   1√ 3  ]

= (1 -  1  3) ÷ (1 +  1  3)

= (1 -  1  3) ÷ (1 +  1  3)

=  3 - 1     3 ÷  3 + 1     3

=  2  3 ÷   4    3

=  2  3 ✕  3  4

=  1  2


Now, Left side
= cos 2θ
= cos 2✕30°
= cos60°
=  1  2   [Since, cos60° =  1  2]

So, Left side = Right side [Showed]



(ii) θ = 30°, show that,

tan 2θ =    2tanθ   1 - tan²θ


Given,
θ = 30°

 
Need to be shown that,

tan 2θ =   2tanθ1 - tan²θ


Here, Right side-

=    2tanθ1 - tan²θ

=    2tan30°1 - tan²30°

= 2tan30° ÷ (1 - tan²30°)

= {2 ✕   1√ 3 } ÷ {1 – (  1√ 3 )²}


=   2√ 3  ÷ (1 –  1  3)


=    2  √ 3  ÷  3 - 1    3

=    2  √ 3  ÷   2    3

=    2√ 3  ✕   3    2

=    3  √ 3 

=   √ 3  ✕ √ 3        √ 3 

= √ 3 


Now, Left side
= tan2θ
= tan(2✕30°)
= tan60°
= √ 3 

So, Left side = Right side   [Showed]

E.6.5  If the elevation angle of the top point of the tree is 60° at a point on the ground 15 m from its base, find the height of the tree.

Solution:

Determine the height of the tree.

According to the question, the figure drawn is,

B is point on the ground and the distance from B to C, BC = 15 meter and

Angle of elevation at point C, ∠ACB = 60°.

Suppose, the height of the tree AB = h meter


We know,

tanθ° = oppositeadjacent

or, tan60° =  AB  BC

or, √ 3  =   h   15

or, h = 15 ✕ √ 3 

∴ h = 25.981 (Approx.)

So, the height of the tree is 25.981 meter. (Approx.)

E.6.6 A ladder with a length of 6 m creates an angle of 60° with the base. What is the height of the roof?

Solution:

Determine the height of the roof.


According to the question, the figure drawn is,

Height of roof from ground, AB = h meter
Length of ground, BC
Length of ladder, AC = 6 meter
Ladder makes 60° angle with ground, ∠ACB = 60°


We know,

sinθ =   opposite hypotenuse

Here, in ΔABC-
sin60° =  AB  AC

or,  √ 3    2 =  h  6   [Since, cos60° =  1  2]

or,  h  6 = √ 3   2

or, h = √ 3  ✕  6  2

or, h = √ 3  ✕ 3

∴ h = 2.5981

So, the height of the roof 2.5981 meter. (Answer)

E.6.7  The elevation angle of the top point of a tower is 60° from a point on the ground. The elevation angle of the tower will be 45° if is 20 m behind from the previous place. What is the height of the roof?

Solution:

Determine the height of the tower.

As per the question we can draw the following diagram.

Here,
Height of tower, AB = h meter
Angle of elevation at point C from ground, ∠ACB = 60°
Angle of elevation at point D from ground, ∠ADB = 45°
Going 20 m back from point C to point D, CD = 20 meter


We know,

tanθ° = oppositeadjacent

Here, in ΔABC-
tan60° =  AB  BC

or, √ 3  =   h BC    [Since, tan60° = √3]

or, h = √ 3  ✕ BC....(i)


Again, ΔABD-এ
tan45° =  AB  BD

or, tan45° =       AB BC + CD  

or, 1 =        h BC + 20   [Since, tan45° = 1]

or, BC + 20 = h

or, h = BC + 20.....(ii)


Therefore, from equations (i) and (ii) we get,

√ 3  ✕ BC = BC + 20

or, √ 3  ✕ BC – BC = 20

or, BC(√ 3  - 1) = 20

or, BC =     20√ 3  - 1

or, BC =         20(1.7321 - 1)   [By using calculator]

or, BC =    200.7321

∴ BC = 27.3187


Now, putting BC = 27.3187, into equation (ii),
    h = BC + 20
or, h = 27.3187 + 20
∴ h = 47.3187 (Approx.)

So, the height of the tower 47.3187 meter. (Approx.) (Answer)

E.6.8 A man standing on the bank of the river observes that on the other side of the river, the elevation angle of the top point of a 100 m tower is 45°. The man starts journey by boat to the tower. But due to water current, the boat reaches the river side with 10 m distance from the tower. Determine the distance from starting point of the man to the ending point.

Solution:

The distance from the man's starting point to his destination has to be determined.

We can draw the following diagram from the question.

Here,
B and C are the two points on the two banks of the given river and the man is standing at point C
Here, BC = River width
Height of the tower, AB = 100 meter
Angle of elevation at point C, ∠ACB = 45°
D is a point on bank 10 m from B where man reaches with boat. Therefore, BD = 10 meter

The distance of the man from the place of departure to the place of destination, DC = ?


We know,

tanθ° = oppositeadjacent

Here, in ΔABC-

tan45° =  AB  BC

or, 1 =  AB  BC   [Since, tan45° = 1]

or, BC = AB

or, BC = 100  [Since, AB= 100 meter]

Here, in ΔBCD-

DC² = BC² + BD²

or, DC² = 100² + 10²

or, DC² = 10000 + 100

or, DC² = 10100

or, DC = √ 10100 

∴ DC = 100.4987 (Approx.)

 
So, the distance from departure place to destination place is 100.4987 meter (Approx.) (Answer)

E.6.9 A man standing on a tower at the bank of the sea observed that a ship was coming towards the port. At that time, the depression angle of the ship was 30°. A few moments later, the depression angle became 45°. If the height of the tower was 50 m, how far did the ship cross during that time?

Solution:

Determine the distance traveled by the ship.

We can draw the following diagram from the question.

Here,
Height of tower, AB = 50 meter
Angle of declination of vessel at poin C, ∠EAC = 30°
Angle of declination of vessel at poin D, ∠EAD = 45°

The distance traveled by the ship, CD = ?

Now, as per the figure,
BC||AE and, AC and AD are common sides
Therefore, ∠EAC = ∠ACB = 30° [Alternate angle]
and, ∠EAD = ∠ADB = 45° [Alternate angle]


We know,

tanθ° = oppositeadjacent

Here, in ΔABC-
tan30° =  AB  BC

or,   1√ 3  =   50  BC   [Since, tan30° =   1√ 3 ]

or,        1 50 ✕ √ 3   =   1  BC

or, 50.√ 3  = BC

or, BC = 50.√ 3 

or, CD + BD = 50.√ 3 .....(i)

Again, in ΔABC-
tan45° =  AB  BD

or, 1 =  50  BD   [Since,tan45°= 1]

or, BD = 50


Now, putting BD = 50, in equation (i),

CD + BD = 50.√ 3 

or, CD + 50 = 50.√ 3 

or, CD = 50.√ 3  - 50

or, CD = 50 ✕ 1.7320 - 50

or, CD = 86.6025 - 50

∴ CD = 36.6025 (Approx.)

So, the distance traveled by the ship is 36.6025 meter (Approx.)

E.6.10 The elevation angle is 45° at a distance of 10 m from your office building. If the elevation angle is θ at a distance of 20 m, what is the value of sinθ and cosθ?

Solution:

Determine the value of sinθ and cosθ.

We can draw the following diagram from the question.

Where,
office building is located at point B,
point C is 10 m from the building, BC = 10 meter
point D is 20 m from the building, BD = 20 meter

Angle of elevation of building at point C is 45°, ∠ACB = 45°
Angle of elevation of building at point D is θ°, ∠ADB = θ°

Sinθ = ? and cosθ = ?


We know,

tanθ° = oppositeadjacent

Here, in ΔABC-
tan45° =  AB  BC

or, 1 =  AB  BC   [Since, tan45° = 1]

or, BC = AB

or, 10 = AB  [Since, BC = 10 meter]

or, AB = 10 meter

Again, ΔABD-এ
tanθ° = ABBD

or, tanθ° =  10  20  [put down the value]

or, tanθ° =  1  2

or,  sinθ°  cosθ° =  1  2   [Since, tanθ° = sinθ°cosθ°]

or, 2sinθ° = cosθ°

or, (2sinθ°)² = (cosθ°)²    [doing square]

or, 4sin²θ° = cos²θ°

or, 4(1 - cos²θ°) = cos²θ°  [Since, sin²θ° + cos²θ° = 1]

or, 4 - 4cos²θ° = cos²θ°

 
or, 4 = cos²θ° + 4cos²θ°

or, 4 = 5cos²θ°

or, 5cos²θ° = 4

or, cos²θ° =  4  5.....(i)

or, cosθ° =  √ 4  √ 5    [doing square root]

∴ cosθ° =   2√ 5 


Again, from equation (ii), we get,

cos²θ° =  4  5

or, 1 - sin²θ° =  4  5   [Since, sin²θ° + cos²θ° = 1]

or, -sin²θ° =  4  5 - 1

or, -sin²θ° = 4 - 5   5

or, -sin²θ° = -  1  5

or, sin²θ° =  1  5

or, sinθ° =   1√ 5   [doing square root]

∴ sinθ° =   1√ 5 


So, sinθ =   1√ 5  and cosθ =   2√ 5     (Answer)