NCERT Class 10 Maths Solutions: Some Applications of Trigonometry
Question:
A statue, 1.6 m tall, stands on a top of pedestal, from a point on the ground, the angle of elevation of the top of statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
Concept in a Minute:
Trigonometry, specifically the tangent ratio (tan θ = opposite/adjacent) in right-angled triangles. We will use angles of elevation to form two right-angled triangles and set up equations using the tangent function.
Explanation:
Let ‘h’ be the height of the pedestal and ‘s’ be the height of the statue.
Let ‘x’ be the distance of the point on the ground from the base of the pedestal.
We have two right-angled triangles:
1. Triangle formed by the point on the ground, the base of the pedestal, and the top of the pedestal.
– Angle of elevation to the top of the pedestal = 45°
– Height (opposite side) = h
– Base (adjacent side) = x
Using tan(45°) = opposite/adjacent, we get tan(45°) = h/x.
Since tan(45°) = 1, we have 1 = h/x, which implies h = x.
2. Triangle formed by the point on the ground, the base of the pedestal, and the top of the statue.
– Angle of elevation to the top of the statue = 60°
– Total height (opposite side) = height of pedestal + height of statue = h + s
– Base (adjacent side) = x
Using tan(60°) = opposite/adjacent, we get tan(60°) = (h + s)/x.
We know s = 1.6 m and tan(60°) = √3.
So, √3 = (h + 1.6)/x.
Now we have a system of two equations:
(i) h = x
(ii) √3 = (h + 1.6)/x
Substitute equation (i) into equation (ii):
√3 = (h + 1.6)/h
Now, solve for ‘h’:
√3 * h = h + 1.6
√3 * h – h = 1.6
h * (√3 – 1) = 1.6
h = 1.6 / (√3 – 1)
To rationalize the denominator, multiply the numerator and denominator by (√3 + 1):
h = 1.6 * (√3 + 1) / ((√3 – 1) * (√3 + 1))
h = 1.6 * (√3 + 1) / (3 – 1)
h = 1.6 * (√3 + 1) / 2
h = 0.8 * (√3 + 1)
Using the approximate value of √3 ≈ 1.732:
h ≈ 0.8 * (1.732 + 1)
h ≈ 0.8 * 2.732
h ≈ 2.1936
The height of the pedestal is 0.8(√3 + 1) meters.
Question:
Two poles of equal heights are standing opposite each other an either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30º, respectively. Find the height of poles and the distance of the point from the poles.
Concept in a Minute:
Trigonometric ratios (tangent) in right-angled triangles.
The sum of angles in a triangle is 180 degrees.
Explanation:
Let the height of the two poles be ‘h’ meters.
Let the distance of the point from the first pole be ‘x’ meters.
Since the road is 80 m wide, the distance of the point from the second pole will be (80 – x) meters.
Consider the right-angled triangle formed by the first pole, the ground, and the line of sight from the point to the top of the first pole.
The angle of elevation is 60°.
Using the tangent ratio:
tan(60°) = Opposite side / Adjacent side
tan(60°) = h / x
√3 = h / x
h = x√3 (Equation 1)
Consider the right-angled triangle formed by the second pole, the ground, and the line of sight from the point to the top of the second pole.
The angle of elevation is 30°.
Using the tangent ratio:
tan(30°) = Opposite side / Adjacent side
tan(30°) = h / (80 – x)
1/√3 = h / (80 – x)
h = (80 – x) / √3 (Equation 2)
Now, we have two expressions for ‘h’. We can equate them to solve for ‘x’.
From Equation 1: x = h / √3
Substitute this value of ‘x’ into Equation 2:
h = (80 – (h / √3)) / √3
h√3 = 80 – h / √3
Multiply the entire equation by √3:
h√3 * √3 = (80 – h / √3) * √3
3h = 80√3 – h
3h + h = 80√3
4h = 80√3
h = (80√3) / 4
h = 20√3 meters
Now, substitute the value of ‘h’ back into Equation 1 to find ‘x’:
h = x√3
20√3 = x√3
x = (20√3) / √3
x = 20 meters
So, the distance of the point from the first pole is 20 meters.
The distance of the point from the second pole is 80 – x = 80 – 20 = 60 meters.
Therefore, the height of the poles is 20√3 meters, and the distances of the point from the poles are 20 meters and 60 meters.
Question:
As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Concept in a Minute:
Trigonometry, specifically the tangent of angles of depression, to form right-angled triangles and solve for unknown distances. Angles of depression from a point above to a point below are equal to the alternate interior angles formed with the horizontal line.
Explanation:
Let the height of the lighthouse be AB = 75 m.
Let the positions of the two ships be C and D, such that ship C is closer to the lighthouse than ship D. Both ships are on the same side of the lighthouse and in a straight line with it.
The angle of depression from the top of the lighthouse (A) to the farther ship (D) is 30°. This means the angle of elevation from ship D to A is also 30° (alternate interior angles). So, angle ADB = 30°.
The angle of depression from the top of the lighthouse (A) to the nearer ship (C) is 45°. This means the angle of elevation from ship C to A is also 45°. So, angle ACB = 45°.
We have two right-angled triangles: triangle ABD and triangle ABC, both with the right angle at B.
In triangle ABD:
tan(ADB) = AB / BD
tan(30°) = 75 / BD
1 / sqrt(3) = 75 / BD
BD = 75 * sqrt(3) m
In triangle ABC:
tan(ACB) = AB / BC
tan(45°) = 75 / BC
1 = 75 / BC
BC = 75 m
The distance between the two ships is the difference between the distances of each ship from the base of the lighthouse, i.e., CD = BD – BC.
CD = (75 * sqrt(3)) – 75
CD = 75(sqrt(3) – 1) m
Substituting the approximate value of sqrt(3) as 1.732:
CD = 75(1.732 – 1)
CD = 75(0.732)
CD = 54.9 m
Therefore, the distance between the two ships is 75(sqrt(3) – 1) meters or approximately 54.9 meters.
Question:
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, where as for the elder children, she wants to have a steep side at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Concept in a Minute:
Trigonometry, specifically the sine function, is used to relate the angle of elevation, the height of a right-angled triangle, and the hypotenuse (which represents the length of the slide). The sine of an angle in a right-angled triangle is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse.
Explanation:
We can model each slide as the hypotenuse of a right-angled triangle. The height of the slide at its top forms the side opposite the angle of inclination to the ground. The length of the slide is the hypotenuse of this triangle.
For the children below 5 years:
Height (opposite side) = 1.5 m
Angle of inclination = 30°
Let the length of the slide be L1.
Using the sine function:
sin(angle) = opposite / hypotenuse
sin(30°) = 1.5 m / L1
We know that sin(30°) = 1/2.
So, 1/2 = 1.5 m / L1
L1 = 2 * 1.5 m
L1 = 3 m
For the elder children:
Height (opposite side) = 3 m
Angle of inclination = 60°
Let the length of the slide be L2.
Using the sine function:
sin(angle) = opposite / hypotenuse
sin(60°) = 3 m / L2
We know that sin(60°) = sqrt(3)/2.
So, sqrt(3)/2 = 3 m / L2
L2 = (2 * 3 m) / sqrt(3)
L2 = 6 m / sqrt(3)
To rationalize the denominator, multiply the numerator and denominator by sqrt(3):
L2 = (6 * sqrt(3)) / (sqrt(3) * sqrt(3))
L2 = 6 * sqrt(3) / 3
L2 = 2 * sqrt(3) m
Therefore, the length of the slide for children below 5 years should be 3 m, and the length of the slide for elder children should be 2 * sqrt(3) m.
Question:
A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
Concept in a Minute:
This problem involves trigonometry, specifically the relationship between angles and sides in a right-angled triangle. We will use trigonometric ratios to find an unknown side when an angle and another side are known. The relevant trigonometric ratio here is sine.
Explanation:
Let’s visualize the situation. The kite is at a height above the ground, the string forms the hypotenuse, and the ground is one of the other sides. This creates a right-angled triangle where:
1. The height of the kite above the ground is the side opposite to the angle of inclination.
2. The length of the string is the hypotenuse of the right-angled triangle.
3. The angle of inclination of the string with the ground is given.
We are given:
Height of the kite (opposite side) = 60 m
Angle of inclination (angle between string and ground) = 60°
We need to find the length of the string (hypotenuse).
The trigonometric ratio that relates the opposite side and the hypotenuse is the sine function:
sin(angle) = Opposite side / Hypotenuse
Let ‘s’ be the length of the string.
So, sin(60°) = 60 m / s
We know that sin(60°) = √3 / 2.
Therefore, √3 / 2 = 60 / s
To find ‘s’, we can rearrange the equation:
s = 60 / (√3 / 2)
s = 60 * (2 / √3)
s = 120 / √3
To rationalize the denominator, we multiply the numerator and denominator by √3:
s = (120 * √3) / (√3 * √3)
s = 120√3 / 3
s = 40√3 m
So, the length of the string is 40√3 meters.
Question:
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Concept in a Minute:
Trigonometric ratios, specifically the tangent function, are used to relate angles of elevation to the sides of a right-angled triangle. In this case, the tower, the ground distance, and the line of sight form a right-angled triangle.
Explanation:
Let the height of the tower be ‘h’ meters.
The distance from the foot of the tower to the point on the ground is given as 30 meters.
The angle of elevation of the top of the tower from this point is 30°.
We can visualize this scenario as a right-angled triangle where:
– The height of the tower is the opposite side to the angle of elevation.
– The distance from the foot of the tower to the point on the ground is the adjacent side to the angle of elevation.
The trigonometric ratio that relates the opposite side and the adjacent side is the tangent (tan).
Therefore, we can write the equation:
tan(angle of elevation) = (opposite side) / (adjacent side)
tan(30°) = h / 30
We know that the value of tan(30°) is 1/√3.
So, the equation becomes:
1/√3 = h / 30
To find the height ‘h’, we can rearrange the equation:
h = 30 * (1/√3)
h = 30 / √3
To rationalize the denominator, we multiply the numerator and the denominator by √3:
h = (30 * √3) / (√3 * √3)
h = 30√3 / 3
h = 10√3
Thus, the height of the tower is 10√3 meters.
To get a numerical value, we can approximate √3 as 1.732:
h ≈ 10 * 1.732
h ≈ 17.32 meters.
The final answer is 10√3 m.
Question:
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower. (Use $\sqrt{3}$ = 1.73)
Concept in a Minute:
This problem involves trigonometry, specifically the use of tangent function in right-angled triangles. We will form two right-angled triangles using the given information, and then use the tangent of the angles of elevation to set up equations relating the unknown height of the tower and the distance from the point on the ground to the building.
Explanation:
Let the height of the building be AB = 20 m.
Let the height of the transmission tower be BC.
Let the total height from the ground to the top of the tower be AC = AB + BC = 20 + BC.
Let the point on the ground be D.
From point D, the angle of elevation of the bottom of the tower (point B) is 45°. This forms a right-angled triangle ABD, where $\angle ADB = 45^\circ$.
In $\triangle ABD$, we have:
tan(45°) = AB / BD
1 = 20 / BD
BD = 20 m
From point D, the angle of elevation of the top of the tower (point C) is 60°. This forms a right-angled triangle ACD, where $\angle ADC = 60^\circ$.
In $\triangle ACD$, we have:
tan(60°) = AC / BD
$\sqrt{3}$ = (20 + BC) / 20
Now, we need to solve for BC, the height of the tower.
20$\sqrt{3}$ = 20 + BC
BC = 20$\sqrt{3}$ – 20
BC = 20($\sqrt{3}$ – 1)
Given that $\sqrt{3}$ = 1.73:
BC = 20(1.73 – 1)
BC = 20(0.73)
BC = 14.6 m
Therefore, the height of the tower is 14.6 m.
Question:
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Concept in a Minute:
Trigonometry (specifically, the tangent of an angle in a right-angled triangle) is used to relate angles of elevation to the heights and distances. Similar triangles or direct trigonometric calculations can be employed.
Explanation:
Let the initial distance of the boy from the building be ‘x’ meters and the distance he walked be ‘d’ meters. The height of the building above the boy’s eye level is (30 – 1.5) = 28.5 m.
When the angle of elevation is 30°, we have a right-angled triangle.
tan(30°) = (Height of building above eye level) / (Initial distance from building)
1/√3 = 28.5 / x
x = 28.5√3 meters
When the angle of elevation is 60°, the boy has walked ‘d’ meters towards the building. The new distance from the building is (x – d) meters.
tan(60°) = (Height of building above eye level) / (Distance from building after walking)
√3 = 28.5 / (x – d)
√3(x – d) = 28.5
x – d = 28.5/√3
d = x – 28.5/√3
Substitute the value of ‘x’ from the first equation into the second equation:
d = 28.5√3 – 28.5/√3
d = 28.5 (√3 – 1/√3)
d = 28.5 ((3 – 1)/√3)
d = 28.5 (2/√3)
d = 57/√3
To rationalize the denominator, multiply the numerator and denominator by √3:
d = (57√3) / 3
d = 19√3 meters
Therefore, the distance he walked towards the building is 19√3 meters.
Next Chapter: Statistics
Refer Some Applications of Trigonometry Notes
Practice Some Applications of Trigonometry Extra Questions
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