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NCERT New Syllabus Mathematics
Class: 10
Exercise 9.1
Topic: Some Applications of Trigonometry
Exploring Real-World Applications of Trigonometry
Trigonometry, often perceived as a collection of abstract mathematical concepts, is deeply intertwined with our daily experiences. In this chapter, we delve into practical scenarios where trigonometric principles are not just applicable but are essential tools. From determining the height of a distant mountain to calculating the optimal angle for a solar panel, trigonometry provides solutions to various real-world problems.
Understanding and applying trigonometric ratios simplifies complex situations, making tasks like measuring distances or angles more manageable. This exploration will enhance your problem-solving skills and deepen your appreciation for the relevance of trigonometry in everyday life.
Key Concepts in Trigonometry: Line of Sight, Angle of Elevation, and Angle of Depression
Line of Sight
The line of sight refers to the imaginary line drawn from the eye of an observer to the point of focus on the object being observed.Angle of Elevation
The angle of elevation is the angle formed between the line of sight and the horizontal when an object is positioned above the observer. This occurs when you raise your head to look at the object.Angle of Depression
The angle of depression is formed between the line of sight and the horizontal when an object is positioned below the observer. This occurs when you lower your head to view the object.Applications of Trigonometry
Trigonometric ratios help determine the height or length of an object or the distance between two distant objects, making them a practical tool in real-world measurements.
EXERCISE 9.1
1. 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° (see Fig).
Solution:
1) PQ represents the vertical pole in this case.2) The length of the rope PR is 20 meters.3) The angle made by the rope with the ground level is 300.4) Now we will determine the height of the vertical pole PQ.sin 30 = PQ/PR1/2 = PQ/2020 x 1/2 = PQPQ = 105) The height of the vertical pole PQ is 10 meters.
2. 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 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.
Solution:
1) The original tree is the sum of the lengths of AB and BC. BC is the length of
the broken part of the tree, and AB is the length of the part of the tree that is still standing.
2) The broken part makes an angle of 300 with the ground.3) The tip of BC, point C, is 8 meters from the base of the tree, point A.
4) The distance from the base of the tree to the tip of the broken part, BC, is
the hypotenuse of a right triangle.
5) Now we will determine the length of the broken tree BC.cos 30 = AC/BC
√3/2 = 8/BC
BC = (8 x 2)/(√3) --------- 1
6) Now we will determine the height of the standing tree AB.
tan 30 = AB/AC
√3/3 = AB/(8)
AB = 8√3/3 --------- 2
7) Height of the tree = AB + BC --------- 3
8) From equations 1, 2 and 3 we have,
Height of the tree = AB + BC
Height of the tree = (8√3/3) + (8 x 2)/(√3)
Height of the tree = (8√3/3)(1 + 2)
Height of the tree = 3(8√3/3)Height of the tree = 8√3
9) The height of the tree is 8√3 m.
3. 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:
1) The smaller slide is 1.5 meters high and is at a 300 angle.
2) Here, in ∆ ABC,
sin 300 = BC/AC
1/2 = 1.5/AC
AC = 1.5 x 2
AC = 3
3) The bigger slide is 3 meters high and is at a 600 angle.
4) Here, in ∆ PQRsin 600 = QR/PR√3/2 = 3/PR
PR = (3 x 2)/√3
PR = (3 x 2 x √3)/(√3 x √3)
PR = (3 x 2 x √3)/(3)
PR = (2 x √3)
5) So,
The length of the slide for the children below the age of 5 years is
3 m. The length of the slide for the elder children is 2√3 m.
4. 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.
Solution:
1) Let the height of the tower be h.
2) The angle of elevation is 300.
3) The observer is at a distance of 30 m from the foot of the tower.
4) Here, in ∆ PQRtan 300 = PQ/QR√3/3 = h/30
h = 30√3/3
h = 10√3 m
5) So, the height of the tower is 10√3 m.
5. 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.
Solution:
1) The kite is flying at a height of 60 m.
2) The inclination of the string with the ground is 60°.
3) Here, in ∆ PQRsin 600 = QR/PR√3/2 = 60/PR
PR = (60 x 2 x √3)/(√3 x √3)
PR = (60 x 2 x √3)/(3)
PR = (20 x 2 x √3)
PR = (40√3)
4) The length of the string PR is 40√3 meters.
6. 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.
Solution:
1) The height of the building AB is 30 m.
2) The boy of height 1.5 m is standing at point D.
3) The angle of elevation from his eyes (C) to the top of the building (B)
increases from 30 degrees to 60 degrees as he walks towards the building.
4) Here
BF = BA – FA
BF = 30 – 1.5
BF = 28.5
5) In ∆ BFC
tan 300 = FB/FC
1/√3 = 28.5/FC
FC = 28.5√3 ---------------- equation 1
6) In ∆ BFEtan 600 = FB/FE√3 = 28.5/FE
FE = 28.5/√3
FE = 28.5√3/3---------------- equation 2
7) In ∆ BFCEC = FC – FE ---------------- equation 3
8) From equations 1, 2 and 3, we have,
EC = FC – FE
EC = (28.5√3) – (28.5√3/3)
EC = 28.5√3(1 – 1/3)
EC = 28.5√3(2/3)EC = 9.5√3(2)EC = 19√3
9) The boy walked 19√3 m towards the building.
7. 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.
Solution:
1) The height of the building PQ is 20 m.2) QR is a tower.3) The angles of elevation of the bottom (Q) and the top (R) of a tower are
45° and 60° respectively.
4) HerePR = PQ + QRPR = 20 + QR ---------------- equation 15) In ∆ SPQtan 450 = PQ/PS
1 = 20/PS
6) In ∆ SPRPS = 20 ---------------- equation 2tan 600 = PR/PS put PS = 20 from equation 2.7) From equations 1, and 3, we have,
√3 = PR/20
PR = 20√3 ---------------- equation 3PR = 20√320 + QR = 20√3QR = 20√3 – 20QR = 20(√3 – 1)8) The height of the tower is 20(√3 – 1) m.
8. A statue, 1.6 m tall stands on the top of a pedestal. From a point on the
ground, the angle of elevation of the top of the 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.
Solution:
1) The height of the statue QR is 1.6 m.2) PQ is a pedestal.3) From a point (S) on the ground, the angle of elevation of the top of the
statue (R) is 60° and from the same point (S), the angle of elevation of the top of the pedestal (Q) is 45°.
4) HerePR = PQ + QRPR = PQ + 1.6 ---------------- equation 15) In ∆ SPQtan 450 = PQ/PS
1 = PQ/PS
6) In ∆ SPRPS = PQ ---------------- equation 2tan 600 = PR/PS
√3 = PR/PS
PR/PS = √3 ---------------- equation 3
7) From equations 1, 2 and 3, we have,PR/PS = √3(PQ + 1.6)/PS = √3(PQ + 1.6)/PQ = √3(PQ + 1.6) = √3 PQ
1.6 = √3 PQ – PQ
1.6 = (√3 – 1) PQ
1.6 = [(√3 – 1)(√3 + 1)] PQ/(√3 + 1)1.6 = [3 – 2] PQ/(√3 + 1)
1.6 = 2 x PQ/(√3 + 1)
[1.6 x (√3 + 1)]/2 = PQPQ = [1.6 x (√3 + 1)]/2
PQ = [0.8 x (√3 + 1)]
8) The height of a pedestal is [0.8 x (√3 + 1)] m.
9. The angle of elevation of the top of a building from the foot of the tower is
30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution:
1) The height of the Tower CD is 50 m.2) Let the height of the building AB be x m.3) The angle of elevation of the top of a building B from the foot of the
tower C is 30° and the angle of elevation of the top of the tower D from the foot of the building A is 60°.
4) In ∆ BACtan 300 = AB/AC
1/√3 = x/AC
AC = √3 x ---------------- equation 15) In ∆ DCAtan 600 = CD/AC
√3 = 50/AC
AC = 50√3/3 ---------------- equation 2
6) From equations 1, and 2, we have,√3 x = 50√3/3
x = 50/3
7) The height of the building is 50/3 m.
10. 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 and the distances of the points from the poles.
Solution:
1) Let the height of the pole AB and CD be x m.2) From a point P between the poles on the road, the angles of elevation of the
top of the poles are 60° and 30°, respectively.
3) Let AP = y, so PC = (80 – y) ---------------- equation 14) In ∆ DCP
tan 300 = DC/PC put DC = x and PC = (80 – y)
1/√3 = x/(80 – y)
(80 – y) = √3 x ---------------- equation 25) In ∆ BPAtan 600 = AB/AP put AB = x and AP = y
√3 = x/y
x = √3 y ---------------- equation 3
6) Put x = √3 y from equations 3 in equation 2, we have,(80 – y) = √3 x(80 – y) = √3 (√3 y)(80 – y) = 3 y4y = 80
y = 20 = AP ---------------- equation 4
7) So CP = (80 – y)
CP = (80 – 20)
CP = 60 ---------------- equation 5
8) From equations 3 and 4, we have
x = √3 y
x = √3 (20)
x = 20√3
9) The height of the pole is 20√3 m. and the point P is at a distance of 20
m from pole AB and 60 m from pole CD.
11. A TV tower stands vertically on the 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° (see the following Fig.). Find the height of the tower and the width of the canal.
Solution:
1) Let the height of the TV Tower AB be h m.2) Let the width of the canal AD be x m. So AC = (x + 20)m3) From a point D on the other bank directly opposite the tower AB, the angle
of elevation of the top of the tower B is 60°.
4) From another point C 20 m away from this point D on the line joing this point
to the foot of the tower, the angle of elevation of the top of the tower B is 30°.
5) In ∆ BAC
tan 300 = AB/AC
1/√3 = h/(x + 20)
(x + 20) = √3 h ---------------- equation 16) In ∆ BADtan 600 = AB/AD
√3 = h/x
h = √3 x ---------------- equation 2
7) Put h = √3 x from equations 2 in equation 1, we have,(x + 20) = √3 h(x + 20) = √3 (√3 x)(x + 20) = 3x2x = 20
x = 10 ---------------- equation 3
8) Put x = 10 from equations 3 in equation 2, we have,h = √3 xh = √3 (10)h = 10√3 ---------------- equation 4
9) The height of the TV Tower is 10√3 m and the width of the canal is
10 m.
12. 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°. Determine the height of the tower.
Solution:
1) Let the height of the Cable Tower CD be h m.
2) So, ED = (h – 7) m. ---------------- equation 1
3) The height of the building AB is 7 m.4) The angle of elevation of the top of a cable tower D is 60° and
the angle of depression of its foot C is 45°.
5) In ∆ CEB
tan 450 = CE/BE
1 = 7/BE
BE = 7 ---------------- equation 26) In ∆ BEDtan 600 = ED/BE
√3 = (h – 7)/BE
(h – 7) = √3 BE ---------------- equation 3
7) Put BE = 7 from equations 2 in equation 3, we have,(h – 7) = √3 BE(h – 7) = √3 (7)h = 7√3 + 7
h = 7(√3 + 1) ---------------- equation 4
8) The height of the Cable Tower CD is 7(√3 + 1) m.
13. 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.
Solution:
1) The height of the lighthouse AB is 75 m.
2) Let the distance between 2 ships C and D be x m.
3) Let the ship D be at distance of y m from the lighthouse AB.4) From the top of the lighthouse to the sea level, the angles of depression
of two ships are 30° and 45°.
5) In ∆ DAB
tan 450 = AB/AD
1 = 75/y
y = 75 ---------------- equation 16) In ∆ BADtan 300 = AB/AC
1/√3 = 75/(x + y)
(x + y) = 75√3 ---------------- equation 2
7) Put y = 75 from equations 1 in equation 2, we have,(x + y) = 75√3(x + 75) = 75√3x = 75√3 – 75
x = 75(√3 – 1) ---------------- equation 3
8) The distance between the two ships is 75(√3 – 1) m.
14. 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° (see the following fig.). Find the distance traveled by the balloon during the interval.
Solution:
1) The first position of the balloon is P and after some time it will reach
to Q.
2) The vertical height of the balloon is BP = CQ = (88.2 – 1.2) = 87 m.
3) Let the ship D be at distance of y m from the lighthouse AB.4) The balloon's angle of elevation from the girl's eyes is 60°.
After some time, the angle of elevation reduces to 30°.
5) The distance travelled by balloon = BC = AC – AB ------------ equation 16) In ∆ ABP
tan 600 = PB/AB
√3 = 87/AB
AB = 87/√3
AB = (87√3)/(√3 x √3)
AB = (87√3)/(3)
AB = 29√3 ---------------- equation 2
7) In ∆ ACQtan 300 = QC/AC
1/√3 = 87/AC
AC = 87√3 ---------------- equation 3
8) From equations 1, 2 and 3, we have,BC = AC – ABBC = 87√3 – 29√3BC = (87 – 29)√3
BC = 58√3 ---------------- equation 3
9) The distance traveled by the balloon during the interval is 58√3 m.
15. A straight highway leads to the foot of a tower. A man standing
at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of 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.
Solution:
1) Let the height of the tower BC be h m.
2) The car traveles from A to C in 6 secs with the uniform speed x m/s.
3) So, the distance AB traveled by the car is 6x m.
4) Let, the distance DB traveled by the car in t sec be tx m.
5) A man standing at the top of the tower observes a car at an angle of
depression of 30°, six seconds later, the angle of depression of the car is found to be 60°.
6) In ∆ ABC
tan 300 = BC/AB
1/√3 = h/(6x + tx)
1/√3 = h/x(6 + t)
x(6 + t) = √3h
h = x(6 + t)/√3 ---------------- equation 1
7) In ∆ DBCtan 600 = BC/BD
√3 = h/tx
h = √3tx ---------------- equation 2
8) From equations 1, and 2, we have,x(6 + t)/√3 = √3tx(6 + t)/√3 = √3t(6 + t) = (√3√3)t
(6 + t) = 3t
3t – t = 6
2t = 6
t = 3 ---------------- equation 3
9) The time taken by the car to reach the foot of the tower is 3 sec.
Unlocking the World with Trigonometry: A Journey Beyond the Angles
As we reach the summit of our trigonometric adventure, one thing becomes crystal clear—trigonometry isn’t just a chapter in a textbook; it’s a key that unlocks the secrets of the world around us. Whether calculating the height of a towering structure, navigating vast oceans, or exploring the cosmos, trigonometry is your guiding light. This mathematical marvel transforms the abstract into the tangible, making the invisible measurable. Keep embracing this journey with curiosity and passion, because the world of angles, triangles, and distances is limitless! Remember, every peak you conquer in mathematics brings you closer to mastering the universe!
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#TrigonometryMagic #MathWizards #Class10NCERT #GeometryOfLife #MathematicsInAction #RealWorldMath #UnlockingTheUniverse #FutureThinkers #STEMPower #ConquerWithMath
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