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For centuries, we have been using trigonometry for the calculation of distances from a point to a far-away object. This phenomenon is moreover used in astronomy, as the objects here are planets that are millions of miles away from our planet. It is also used in navigating geographical locations and constructing various maps. Theodolite is an impeccable example of its application, it is surveying instruments, which is based on the principles of trigonometry, especially for measuring angles with a rotating telescope.
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EXPLANATION
[Click Here for Sample Questions]
Heights and distances are measured through trigonometry.
The line of sight is the line drawn from the eye of the observer to the point on the object viewed by the observer.
The angle of elevation of the point viewed is above the horizontal level i.e. the case when we raise our head to look at the object.
The angle of depression of a point on the object being viewed is the angle formed by the line of sight with the horizontal when the point is below the horizontal level, i.e. the case when we lower our head to look at the point being viewed.
The questions mentioned below are examples or samples of the various angles and lengths mentioned above. The following table shows the values of the corresponding angles with respect to trigonometry:
| θ | 0 | 30 | 45 | 60 | 90 |
| Sin | 0 | 1/2 | 1/√2 | √3/2 | 1 |
| Cos | 1 | √3/2 | 1/√2 | 1/2 | 0 |
| Tan | 0 | 1/√3 | 1 | √3 | Not defined |
| Cot | Not defined | √3 | 1 | 1/√3 | 0 |
| Sec | 1 | 2/√3 | √2 | 2 | Not defined |
| Cosec | Not defined | 2 | √2 | 2/√3 | 1 |
PROBLEMS
Q1) A ladder 15 m long just reaches the top of a vertical wall. If the ladder AC makes an angle of 60? with the wall, then calculate the height of the wall AB. (3 marks)
Solution:
Given: Angle ABC = 90?
Angle BAC = 60?
AC = 15 m
If angles ABC and BAC are 90? and 60? respectively, then the angle ACB is 30?.
Sin30 = AB/AC
1/2 =AB/15
AB = 15/2
AB = 7.5 m
Therefore, the height of the wall AB is 7.5 m.
Q2) The angles of elevation on the top of a tower from two points at a distance of 4 m and 9m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6m. (4 marks)
Solution:
Let AB be the tower,
C and D be the two points with distance 4m and 9m from the base respectively.
A/q
In the right triangle ABC,
tanX = AB/BC
tanX = AB/4
AB = 4tanX ………..(1)
Also,
In right triangle ABD,
tan (90? – X) = AB/BD
cotX = AB/BD
AB = 9 cot X………(2)
Multiplying equation 1 and 2
AB2 = 9 cot X × 4 tan X
AB2 = 36
AB = 6
Hence, the height of the tower is 6 m.
Q3) From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60? and the angle of depression of its foot is 45?. Find out the height of the tower. (3 marks)
Solution:
Let AB be the building of height 7 m and EC be the height of a tower. A is the point from where elevation of the tower is 60? and the angle of depression of its foot is 45?
EC = DE + CD
Also, CD = AB = 7 m
And, BC = AD
A/q
In right triangle ABC,
tan 45? = AB/BC
1 = 7/BC
BC = 7 m= AD
Also, In right triangle ADE,
tan 60? = DE/AD
√3 = DE/7
DE = 7/√3 m
Height of the tower = EC = DE + CD
= (7√3 + 7) m
= 7 (√3 + 1) m
Therefore, the height of the tower is 7(√3 + 1) m.
Q4) The shadow of a tower standing on level ground is found to be 40m longer when the sun’s altitude is 30? than when it is 60?. Find out the height of the tower. (4 marks)
Solution:
Let AB be the tower and BC be the length of its shadow when the sun’s altitude is 60? and DB be the length of the shadow when the angle of elevation 30?
Let’s assume, AB= h m and BC = x m
DB = (40 + X) m
In right triangle ABC,
tan 60? = AB/BC
√3 = h/x ……….(1)
In right triangle ABD,
tan 30? = AB/BD
1/√3 = h/(x + 40) ……….(2)
From 1 and 2
X (√3) (√3) = X + 40
3X = X + 40
2X = 40
X = 20
Substituting, X = 20 in 1
h = 20√3
Therefore, the height of the tower is 20√3 m.
Q5) A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 30? with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Determine the height of the tree. (3 marks)
Solution:
Using given instructions, draw a figure. Let AC be the broken part of the tree. Angle C = 30?
BC = 8m
Total height of the tree is the sum of AB and AC i.e. AB+AC,
In right triangle ABC
Using Cosine and tangent angles,
Cos 30? = √3/2
√3/2 = 8/AC
AB = 8√3 ……(2)
From 1 and 2
Total height of the tree = AB + AC = 16/√3 + 8/√3 = 24/√3= 8√3 m.
Q6) A pole 6m high casts a shadow 2/√3 m long on the ground. What would be the sun’s elevation? (3 marks)
Solution:
Let BC be the pole and AB be its shadow.
In triangle ABC,
tan CAB = BC/AB
= 6/2√3
=3/√3
= √3
tan CAB = tan 60?
CAB = 60?
Therefore, the sun’s elevation is 60?
Q7) The angle of elevation of the top of a tower from two points distance s and t from its foot are complementary. Prove that the height of the tower is √st (3 marks)
Solution:
Let BC = s ; PC = t
Let the height of the tower be AB = h
ABC = θ and APC = 90? – θ
(Because, the angle of elevation of the top of the tower from two points P and B are complementary)
In triangle ABC,
tan θ = AC/BC = h/s ……(1)
tan (90? – θ) = AC/PC = h/t
Cot θ = h/t ………..(2)
Multiplying 1 and 2,
Tan θ × cotθ = (h/s) × (h/t)
1 = h2/st
h2 = st
h = √st
Therefore, the height of the tower is √st
Q8) A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant traveled by the balloon during the interval. (3 marks)
Solution:
Let the initial Position of the balloon be A and the final position be B. Height of the balloon above the girl height = 88.2 m – 1.2 m = 87 m
Distance travelled by the balloon = DE = CE – CD
A/q
In right triangle BEC,
Tan 30? = BE/CE
1/√3 = 87/CE
CE = 87√3 m
Also,
In right triangle ADC,
Tan 60? = AD/CD
√3 = 87/CD
CD = 87/√3 = 29/√3 m
Distance traveled by the balloon = DE= CE – CD = (87√3 – 29√3) m
= 29√3 (3-1)
= 58√3 m
Therefore, the distance travelled by the balloon during the interval is 58√3 m.
Q9) A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at the angle of depression of 30?, which is approaching the foot of the tower with a uniform speed. Six Seconds later, the angle of the depression of the car is found to be 60?. Find the time taken by the car to reach the foot of the tower from this point. (4 marks)
Solution:
Let AB be the tower,
D is the initial and final position of the car respectively.
Angles of depression are measured from A.
BC is the distance from the foot of the tower to the car.
A/q,
In right triangle ABC,
tan 60? = AB/BC
√3 = AB/BC
BC = AB/√3
Also,
In right triangle ABD,
tan 30? = AB/BD
1/√3 = AB/ (BC+CD)
AB√3 = BC + CD
AB√3 = AB√3 + CD
CD = AB√3 + CD
CD = AB (√3 – 1/√3)
CD = 2 AB/ √3
The distance of BC is half of CD. Therefore, the time taken is also half.
Time taken by the car to travel distance CD = 6 seconds.
Time taken by car to travel BC = 6/2 = 3 seconds.
Q10) The angles of elevation of the top of the tower from two points at a distance of 4m and 9m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6m. (3 marks)
Solution:
Let AB be the tower,
C and D be the two points with a distance of 4 m and 9 m from the base respectively.
A/q
In the right triangle ABC,
tan X = AB/BC
tan X = AB/4
AB = 4 tan X ……………(1)
Also,
In the right triangle ABD,
tan (90? – X) = AB/BD
cot X = AB/9
AB = 9 cot X …………(2)
Multiplying equation 1 and 2
We get,
AB2 = 36
AB = 6
The height of the tower is 6m.
Hence, Proved.
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