Some Applications of Trigonometry — Class 10 Maths Notes

Some Applications of Trigonometry · Class 10 Maths · 19 topics.

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Topics covered in Some Applications of Trigonometry

  1. 1.Introduction of Some Applications of Trignometry

    1. Measuring Heights: Trigonometry helps in finding the height of tall objects like buildings, mountains, and trees when you can't measure them directly.

    2. Navigation: Sailors and pilots use trigonometry to find their direction and distance from a particular point.

    3. Engineering: Bridges, buildings, and machines often require trigonometry for their design and construction.

    4. Physics: Trigonometry is used in calculating forces, energy, and waves.

    5. Astronomy: Scientists use trigonometry to calculate distances between stars and planets.

      Real-Life Example:

      Suppose you want to find the height of a tree, and you know the angle of elevation when you look at the top of the tree is 30 degrees. You are standing 20 meters away from the tree.

      To find the height (h) of the tree, you can use the formula:

      tan⁡(angle)=Height of treeDistance from treetan(angle)=Distance from treeHeight of tree​

      So,

      tan⁡(30)=ℎ20tan(30)=20h​ℎ=20×tan⁡(30)=20×13≈11.54 metersh=20×tan(30)=20×3​1​≈11.54 meters

      So, the height of the tree is approximately 11.54 meters.

  2. 2.Height & distance

    The concept of "Height and Distance" in trigonometry is often used to find the height of an object or the distance to an object when you can't measure it directly. The most commonly used trigonometric ratios for this are sine, cosine, and tangent.

    Simple Explanation

    1. Angle of Elevation: The angle you look up at to see the top of an object.
    2. Angle of Depression: The angle you look down at to see the bottom of an object.
    3. Adjacent Side: The side that is next to the angle you are considering.
    4. Opposite Side: The side that is opposite to the angle you are considering.
    5. Hypotenuse: The longest side in a right triangle, opposite to the right angle.

      Real-Life Example:

      Suppose you want to find the height of a building and you are standing 50 meters away from it. If the angle of elevation when you look at the top of the building is 60 degrees, you can use the formula:

      tan⁡(angle)=Height of buildingDistance from buildingtan(angle)=Distance from buildingHeight of building​

      So,

      tan⁡(60)=ℎ50tan(60)=50h​ℎ=50×tan⁡(60)=50×3≈86.6 metersh=50×tan(60)=50×3​≈86.6 meters

  3. 3.A tower stands vertically on the ground. From a point on the ground, which is 30 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.

    To find the height of the tower, you can use trigonometry. Specifically, you can use the tangent of the angle of elevation, which in this case is 60°. The formula for tangent in terms of opposite (height of the tower, which we'll call ℎh) and adjacent sides (distance from the point to the foot of the tower, which is 30 m) is:

    tan⁡(angle)=Opposite sideAdjacent sidetan(angle)=Adjacent sideOpposite side​

    So, for this problem:

    tan⁡(60)=ℎ30tan(60)=30h​

    First, we find tan⁡(60)tan(60), which is 33​.

    3=ℎ303​=30h​

    Now, solve for ℎh:

    ℎ=30×3≈30×1.732≈51.96 metersh=30×3​≈30×1.732≈51.96 meters

    So, the height of the tower is approximately 51.96 meters.

  4. 4.An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3m below the top of the pole to undertake the repair work . What should be the length of the ladder that she should use which, when ......

    An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3m below the top of the pole to undertake the repair work. What should be the length of the ladder that she should use which, when inclined at an angle of 60° to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? Finding the Length of the Ladder

    The electrician needs to reach a point 1.3 m below the top of the 5 m pole. So, the height she needs to reach is 5−1.3=3.75m−1.3m=3.7m.

    The ladder makes an angle of 60° with the ground. To find the length of the ladder (L), we can use the sine function:

    sin⁡(60)=Opposite side (height)Hypotenuse (length of ladder)sin(60)=Hypotenuse (length of ladder)Opposite side (height)​sin⁡(60)=3.7sin(60)=L3.7​

    We know that sin⁡(60)=32sin(60)=23​​.

    32=3.723​​=L3.7​

    Solving for L:

    =2×3.73≈7.41.732≈4.27 metersL=3​2×3.7​≈1.7327.4​≈4.27 meters

    Finding the Distance from the Foot of the Pole

    To find how far from the foot of the pole she should place the ladder, we can use the cosine function:

    cos⁡(60)=Adjacent side (distance from pole)Hypotenuse (length of ladder)cos(60)=Hypotenuse (length of ladder)Adjacent side (distance from pole)​cos⁡(60)=4.27cos(60)=4.27d​

    We know that cos⁡(60)=12cos(60)=21​.

    12=4.2721​=4.27d​

    Solving for d:

    =4.272=2.135 metersd=24.27​=2.135 meters

  5. 5.An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney?

    Finding the Height of the Chimney

    The observer is 1.5 m tall and is standing 28.5 m away from the chimney. The angle of elevation from her eyes to the top of the chimney is 45°.

    First, let's find the height from the observer's eyes to the top of the chimney. For a 45° angle, tan⁡(45)=1tan(45)=1.

    tan⁡(45)=Height from eyes to top of chimneyDistance from chimneytan(45)=Distance from chimneyHeight from eyes to top of chimney​1=ℎ28.51=28.5h​

    Solving for ℎh:

    ℎ=28.5 metersh=28.5 meters

    Now, the observer is 1.5 m tall, and the height from her eyes to the top of the chimney is 28.5 m. Therefore, the total height of the chimney is 28.5+1.5=3028.5m+1.5m=30m.

  6. 6.The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.

    To find the height of the tower, we can use trigonometry, specifically the tangent function. The tangent of an angle in a right triangle is the opposite side (height of the tower, which we'll call ℎh) divided by the adjacent side (length of the shadow).

    When the Sun's altitude is 30°

    Let's say the length of the shadow when the Sun's altitude is 30° is x.

    tan⁡(30)=ℎtan(30)=xh​

    We know tan⁡(30)=13tan(30)=3​1​.

    13=ℎ3​1​=xh​=ℎ3x=h3​

    When the Sun's altitude is 60°

    The shadow is 40 m longer when the Sun's altitude is 30° than when it is 60°. So, the length of the shadow when the Sun's altitude is 60° is −40x−40.

    tan⁡(60)=ℎ−40tan(60)=x−40h​

    We know tan⁡(60)=3tan(60)=3​.

    3=ℎ−403​=x−40h​−40=ℎ3x−40=3​h​

    Solving for ℎh

    We have two equations:

    1. =ℎ3x=h3​
    2. −40=ℎ3x−40=3​h​

    Substitute the value of x from the first equation into the second equation:

    ℎ3−40=ℎ3h3​−40=3​h​

    Multiply by 33​:

    ℎ×3−403=ℎh×3−403​=h3ℎ−ℎ=4033h−h=403​2ℎ=4032h=403​ℎ=203≈34.64 metersh=203​≈34.64 meters

    So, the height of the tower is approximately 34.64 meters.

  7. 7.The angles of depression of the top and the bottom of an 16 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multistoreyed building and the distance between the two buildings.

    To find the height of the multi-storeyed building and the distance between the two buildings, we can use trigonometry. Specifically, we'll use the tangent function, which is the ratio of the opposite side to the adjacent side in a right triangle.

    Finding the Height of the Multi-Storeyed Building

    Let's say the height of the multi-storeyed building is H meters, and the distance between the two buildings is D meters.

    1. For the angle of depression of 45° to the bottom of the 16 m building:
    tan⁡(45)=tan(45)=DH​

    Since tan⁡(45)=1tan(45)=1,

    =(Equation 1)H=D(Equation 1)
    1. For the angle of depression of 30° to the top of the 16 m building:
    tan⁡(30)=−16tan(30)=DH−16​

    We know tan⁡(30)=13tan(30)=3​1​,

    13=−163​1​=DH−16​=(−16)3(Equation 2)D=(H−16)3​(Equation 2)

    Solving for H and D

    We have two equations:

    1. =H=D
    2. =(−16)3D=(H−16)3​

    Substitute the value of D from Equation 1 into Equation 2:

    =(−16)3H=(H−16)3​=3−163H=H3​−163​−3=−163H−H3​=−163​(1−3)=−163H(1−3​)=−163​=−1631−3H=1−3​−163​​=−163−3+1H=−3​+1−163​​=1633−1H=3​−1163​​=163×3+13−1H=163​×3−13​+1​=8×(3+1)H=8×(3​+1)=83+8H=83​+8=8(3+1)H=8(3​+1)=8×2.732+8H=8×2.732+8=21.856+8H=21.856+8=29.856 meters (approximately)H=29.856 meters (approximately)

    Using Equation 1, =D=H,

    =29.856 meters (approximately)D=29.856 meters (approximately)

  8. 8.From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30° and 45°, respectively. If the bridge is at a height of 3 m from the banks, find the width of the river

    To find the width of the river, we can use trigonometry, specifically the tangent function. The tangent of an angle in a right triangle is the opposite side (height of the bridge, which is 3 m) divided by the adjacent side (distance to the bank).

    Finding the Width of the River

    Let's say the distance from the point on the bridge to the first bank is 1D1​ meters, and the distance to the second bank is 2D2​ meters.

    1. For the angle of depression of 30° to the first bank:
    tan⁡(30)=31tan(30)=D1​3​

    We know tan⁡(30)=13tan(30)=3​1​,

    13=313​1​=D1​3​1=33(Equation 1)D1​=33​(Equation 1)
    1. For the angle of depression of 45° to the second bank:
    tan⁡(45)=32tan(45)=D2​3​

    Since tan⁡(45)=1tan(45)=1,

    2=3(Equation 2)D2​=3(Equation 2)

    Finding the Total Width of the River

    The total width of the river would be 1+2D1​+D2​.

    1+2=33+3D1​+D2​=33​+31+2=3(3+1)D1​+D2​=3(3​+1)1+2=3×2.732+3D1​+D2​=3×2.732+31+2=8.196+3D1​+D2​=8.196+31+2=11.196 meters (approximately)D1​+D2​=11.196 meters (approximately)

  9. 9.A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30 degree.

    In this problem, we have a rope that is 20 meters long. This rope is stretched from the top of a vertical pole to the ground, making an angle of 30° with the ground. We are interested in finding the height of this pole.

    To find the height, we can use trigonometry, specifically the sine function. In a right triangle, the sine of an angle is defined as the length of the opposite side divided by the length of the hypotenuse.

    The formula for sine in terms of the height (ℎh) of the pole and the length of the rope (20 m) is:

    sin⁡(30)=ℎ20sin(30)=20h​

    We know that the sine of 30 degrees is 1221​. Plugging this into our equation:

    12=ℎ2021​=20h​

    Simplifying, we find:

    ℎ=10 metersh=10 meters

    So, the height of the pole is 10 meters.

  10. 10.A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

    To find the height of the tree, we can use trigonometry, specifically the sine function. The sine of an angle in a right triangle is the opposite side (height of the tree, which we'll call ℎh) divided by the hypotenuse (length of the broken part of the tree, which is 8 m).

    The formula for sine in terms of the height (ℎh) of the tree and the length of the broken part (8 m) is:

    sin⁡(30)=ℎ8sin(30)=8h​

    We know that the sine of 30 degrees is 1221​. Plugging this into our equation:

    12=ℎ821​=8h​

    Simplifying, we find:

    ℎ=4 metersh=4 meters

    So, the height of the tree is 4 meters.

  11. 11.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, whereas for elder.....

    Que- 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, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case? Solution To find the length of each slide, we can use trigonometry, specifically the cosine function. The cosine of an angle in a right triangle is the adjacent side (height of the slide) divided by the hypotenuse (length of the slide).

    For Children Below the Age of 5 Years:

    1. Height of the slide: 1.5 m
    2. Angle of inclination: 30°

    The formula for cosine is:

    cos⁡(30)=1.5Length of slide (let’s call it 1)cos(30)=Length of slide (let’s call it L1​)1.5​

    We know cos⁡(30)=32cos(30)=23​​,

    32=1.5123​​=L1​1.5​

    Solving for 1L1​:

    1=1.5×23L1​=3​1.5×2​1=33L1​=3​3​1=3≈1.732 metersL1​=3​≈1.732 meters

    For Elder Children:

    1. Height of the slide: 3 m
    2. Angle of inclination: 60°

    The formula for cosine is:

    cos⁡(60)=3Length of slide (let’s call it 2)cos(60)=Length of slide (let’s call it L2​)3​

    We know cos⁡(60)=12cos(60)=21​,

    12=3221​=L2​3​

    Solving for 2L2​:

    2=3×21L2​=13×2​2=6 metersL2​=6 meters

  12. 12.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.

    Problem Statement:

    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°. We need to find the height of the tower.

    Concept:

    We'll use trigonometry to solve this problem, specifically the tangent function. In a right triangle, the tangent of an angle is the opposite side divided by the adjacent side.

    Mathematical Solution:

    The formula for tangent in this context is:

    tan⁡(30)=Height of the tower (let’s call it ℎ)30tan(30)=30Height of the tower (let’s call it h)​

    We know tan⁡(30)=13tan(30)=3​1​,

    13=ℎ303​1​=30h​

    Solving for ℎh:

    ℎ=303h=3​30​ℎ=103≈17.32 metersh=103​≈17.32 meters

  13. 13.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 .......

    Que - 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. Problem Statement:

    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°. We need to find the length of the string.

    Concept:

    We'll use trigonometry to solve this problem, specifically the sine function. In a right triangle, the sine of an angle is the opposite side divided by the hypotenuse.

    Mathematical Solution:

    The formula for sine in this context is:

    sin⁡(60)=Height of the kite (60 m)Length of the string (let’s call it)sin(60)=Length of the string (let’s call it L)Height of the kite (60 m)​

    We know sin⁡(60)=32sin(60)=23​​,

    32=6023​​=L60​

    Solving for L:

    =60×23L=3​60×2​=1203L=3​120​=403≈69.28 metersL=403​≈69.28 meters

    So, the length of the string is approximately 69.28 meters.

  14. 14.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.

    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. We need to find the distance he walked towards the building.

    Concept:

    We'll use trigonometry to solve this problem, specifically the tangent function. In a right triangle, the tangent of an angle is the opposite side divided by the adjacent side.

    Mathematical Solution:

    Let's consider two scenarios:

    1. When the angle is 30°:
      The height of the building is 30 m, and the boy is 1.5 m tall. So, the effective height to consider is 30−1.5=28.530−1.5=28.5 m.
      Let the initial distance from the boy to the building be 1d1​.
    tan⁡(30)=28.51tan(30)=d1​28.5​13=28.513​1​=d1​28.5​1=28.5×3≈49.35 metersd1​=28.5×3​≈49.35 meters
    1. When the angle is 60°:
      The effective height remains 28.528.5 m.
      Let the final distance from the boy to the building be 2d2​.
    tan⁡(60)=28.52tan(60)=d2​28.5​3=28.523​=d2​28.5​2=28.53≈16.46 metersd2​=3​28.5​≈16.46 meters

    Finally, the distance he walked towards the building is 1−2=49.35−16.46≈32.89d1​−d2​=49.35−16.46≈32.89 meters.

  15. 15.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.

    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. We need to find the height of the tower.

    Concept:

    We'll use trigonometry to solve this problem, specifically the tangent function. In a right triangle, the tangent of an angle is the opposite side divided by the adjacent side.

    Mathematical Solution:

    Let the distance from the point on the ground to the base of the building be d.

    1. When the angle is 45° for the bottom of the building:
      The height of the building is 20 m.
    tan⁡(45)=20tan(45)=d20​1=201=d20​=20 metersd=20 meters
    1. When the angle is 60° for the top of the tower:
      Let the height of the tower be ℎh.
      The effective height from the ground to the top of the tower is 20+ℎ20+h.
    tan⁡(60)=20+ℎtan(60)=d20+h​3=20+ℎ203​=2020+h​203=20+ℎ203​=20+hℎ=203−20h=203​−20ℎ=20(3−1)h=20(3​−1)ℎ≈20(1.732−1)h≈20(1.732−1)ℎ≈20×0.732h≈20×0.732ℎ≈14.64 metersh≈14.64 meters

    So, the height of the tower is approximately 14.64 meters.

  16. 16.Two poles of equal heights are standing opposite each other on 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 30°, respectively. Find the height of the poles

    Two poles of equal heights are standing opposite each other on 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 30°, respectively. We need to find the height of the poles and the distances of the point from the poles.

    Concept:

    We'll use trigonometry to solve this problem, specifically the tangent function. In a right triangle, the tangent of an angle is the opposite side divided by the adjacent side.

    Mathematical Solution:

    Let the height of the poles be ℎh.

    1. When the angle is 60° for one pole:
      Let the distance from the point to this pole be 1d1​.
    tan⁡(60)=ℎ1tan(60)=d1​h​3=ℎ13​=d1​h​1=ℎ3d1​=3​h​
    1. When the angle is 30° for the other pole:
      Let the distance from the point to this pole be 2d2​.
    tan⁡(30)=ℎ2tan(30)=d2​h​13=ℎ23​1​=d2​h​2=ℎ3d2​=h3​

    The total distance between the poles is 80 m, so:

    1+2=80d1​+d2​=80ℎ3+ℎ3=803​h​+h3​=80ℎ(13+3)=80h(3​1​+3​)=80ℎ(3+33)=80h(3​3​+3​)=80ℎ=8034h=4803​​ℎ=203≈34.64 metersh=203​≈34.64 meters

    And 1=34.643≈20d1​=3​34.64​≈20 meters, 2=34.643≈60d2​=34.643​≈60 meters.

  17. 17.A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joing this point to the foot of

    Que- A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal. Let the height of the tower be

    ℎh and the width of the canal be w.

    1. When the angle is 60°:
    tan⁡(60)=ℎtan(60)=wh​3=ℎ3​=wh​ℎ=3h=w3​
    1. When the angle is 30°:
      The distance from the new point to the foot of the tower is +20w+20 m.
    tan⁡(30)=ℎ+20tan(30)=w+20h​13=ℎ+203​1​=w+20h​ℎ=(+20)13h=(w+20)3​1​

    Equating the two equations for ℎh:

    3=(+20)13w3​=(w+20)3​1​=203−13w=3​−3​1​20​=2032w=2203​​=103≈17.32 metersw=103​≈17.32 metersℎ=103×3=30 metersh=103​×3​=30 meters

  18. 18.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 is 60°. After some time, the angle of elevation reduces to..

    Que. 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 is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval. Solution Step 1: Calculate the Height from the Girl's Eyes to the Balloon

    The height of the balloon from the ground is 88.2 m, and the girl is 1.2 m tall. So, the height from the girl's eyes to the balloon is 88.2 m−1.2 m=87 m88.2m−1.2m=87m.

    Step 2: Find the Initial Distance (1D1​) when the angle is 60°

    Using the tangent function,

    tan⁡(60)=871tan(60)=D1​87​3=8713​=D1​87​1=873≈50.2 mD1​=3​87​≈50.2m

    Step 3: Find the Final Distance (2D2​) when the angle is 30°

    tan⁡(30)=872tan(30)=D2​87​13=8723​1​=D2​87​2=873≈150.6 mD2​=873​≈150.6m

    Step 4: Find the Distance Travelled by the Balloon

    Distance travelled=2−1=150.6 m−50.2 m=100.4 mDistance travelled=D2​−D1​=150.6m−50.2m=100.4m

  19. 19.Quick Revision

    1. Introduction of Some Applications of Trigonometry:

    Trigonometry is not just about solving problems on paper; it has real-world applications too. It’s used in navigation to find the distance of the shore from a point in the sea, in astronomy to calculate distances between celestial bodies, and in physics to analyze wave motion. Architects use it to calculate the slope of roofs and engineers to determine the forces on bridges.

    2. Height and Distance: One of the most common applications of trigonometry is to find the height of an object or the distance from an object when it's not possible to measure it directly. For example, you can calculate the height of a mountain by measuring the angle of elevation from a certain point.

    Important Formulas:

    • Angle of Elevation: The angle above horizontal your line of sight is when looking at the top of an object.
    • Angle of Depression: The angle below horizontal your line of sight is when looking down to the base of an object from a height.
    • To find the height (h) of an object when the angle of elevation (θ) and the distance (d) from the object are known, use the formula: ℎ=⋅tan⁡h=d⋅tan(θ).
    • To find the distance (d) from an object when the height (h) and the angle of elevation (θ) are known, use the formula: =ℎ⋅cot⁡d=h⋅cot(θ).

    Implementing These Formulas: Suppose you want to find the height of a tree. You step back 30 meters and measure the angle of elevation with a clinometer and find it to be 45°. Using the tangent ratio, you calculate the height by multiplying the distance from the tree (30 meters) by the tangent of the angle (1 for 45°), which gives you the height of the tree as 30 meters.

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