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Some Application of Trigonometry Ex- 9.1

 Hi Friends and Champs

Before we proceed to the next exercise, let us understand some points.

Let  ΔABC is a right-angle triangle right angle at B. Draw a line "l" from C parallel to line AB like in below given figure.

then,
line of sight : is the straight line , where observer is at one side and the object is at other side . In the above given figure the line of sight is the line "AC", where observer could be at any one end of line AC and object is at other hand.
Case - I  
When observer is at end A and object is at end C.
when object is above the horizontal line , i.e. the observer look at the object by raising own head , then the angle between the line of sight and horizontal line ( angle between AC and AB ) is called Angle of elevation.
Case - II 
When observer is at end C and object is at end A.
when object is below the horizontal line , i.e. the observer look at the object by lowering own head ,then the angle between the line of sight and horizontal line ( angle between line "l" and AC ) is called Angle of depression.
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. 9.11).
Solution:
Since tan θ = P /H= AB / AC therefore
sin30° = AB / 20
=> 1 / 2 = AB / 20
=> AB = 20 / 2
=> AB =10 Ans.
2. 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. 
Solution:

In above figure BC = 8 m ( given ) and <C = 30°
Since tanθ = AB / BC = P / B
=> tan 30° = AB / 8
=>AB = 8 x tan 30°
=>AB = 8 x 1/√3
=>AB = 8√3 height of the tree Ans.
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:
Let Fig-1 is for children below 5 years ,where
<C = 30° , AB = 1.5 m , AC = ?
and Fig-2 is for children above 5 years, where
<Q = 60° , PR = 3 m , PQ = ?
In Fig - 1
sin 30° = AB / AC
=> 1/2 = 1.5 / AC
=> AC =3 m  Ans. length of slide for children below 5 year
In Fig - 2
Sin60° = PR / PQ
=>√3/2 = 3/PQ
=>PQ =6/√3
=>PQ=6√3/√3√3  multiple and divide by √3
=>PQ=6√3/3
=>PQ=2√3  Ans. length of slide for children above 5 year
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:
BC = 30 mtr ( given )
<C = 30°
tan30° = AB / 30
=> AB = 30 x 1/√3
=>AB = 30√3 / 3         multiple and divide by √3
=>AB =10√3 =            height of the tower. Ans

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:

AB = 60 m  ( given )

AC = ?

< C = 60°

Since , Sinθ = P / H

=>Sin60° = AB / AC

=>√3/2 = 60 / AC

=>AC = 2 x 60 /√3

=>AC = 120 x √3 / √3 x √3   ( divide and multiply by √3 )

=>AC = 120x√3 /3

=>AC = 40√3

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: 

Let DF = 1.5 m is the boy standing at some distance ED from a building AB =30 m , need to find the distance CD =?
AE = AB - EB
AE = 30 - 1.5                since EB =DF=1.5 m
AE = 28.5
tan D = AE/ED
tan 30° = 28.5/ED
1/√3 = 28.5/ED
ED = 28.5√3
tan C = AE/EC
tan 60° = 28.5/EC
√3 = 28.5/EC
EC = 28.5/√3
since CD = ED - EC
=>CD = 28.5√3 - 28.5/√3
=>CD = 28.5(√3 - 1/√3)
=>CD = 57/√3
=>CD = 57√3 /3            multiply and divide by √3
=>CD = 19√3

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:
BC = 20 m ( given)
< ADB = 60° (given)
<BDC =45° (given)
AC=?
tan45° =BC/CD
1=20/CD
=>CD=20 m
tan 60° = AC/CD
=>AC=CD x tan60°
=>AC=20 x √3
=>AC = 20√3 
Hence tower of height = AC - BC = 20√3 - 20=20(√3 - 1) m  Ans.  

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:
AB = Height of Statue = 1.6 m
<ADC = 60° = Angle of the top of the statue
<BDC = 45° =Angle of the top of the pedestal
BC = ?
tan 45° = BC/CD
1 =BC/CD
=>BC =CD
tan 60° =  AC/CD
√3=(AB+BC)/BC        since BC=CD and AC = AB + BC
√3=(1.6 + BC)/BC
√3BC = 1.6 + BC
(√3 - 1) BC = 1.6
BC = 1.6 / (√3 - 1)
BC = 1.6(√3 + 1)/(√3 - 1)(√3 + 1)    multiply and divide by (√3 + 1)
BC = 1.6(√3 + 1)/2
BC = 0.8(√3 + 1) m = Height of the pedestal Ans.
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:
AC = 50 m Height of the tower (given)
ED = ? height of the building
<D = 60° degree Angle of elevation from the foot of the building to the tower
<C = 30° degree Angle of elevation from the foot of the tower to the building
tan 30° = ED / CD
=>CD = ED/tan 30° -----(1)
tan 60° = AC / CD
=>CD = 50/tan60° ------(2)
From eq (1) and (2)
ED/tan30° = 50 / tan60°
ED = 50 x tan30° / tan60°
ED = 50 x (1/ √3) / √3
ED = 50 / 3
ED = 16 2/3 =Height of the Building  Ans.
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 point from the poles.
Solution:
DC = 80 m = road breadth
AC = ED = ? Height of Poles
CF =? , FD = ?
tan 60° = AC/CF -------- (1)
tan 30° = ED/FD
tan 30° = AC/FD --------(2)
from eq (1) and (2)
CF tan 60° = FD tan30°
FD / CF = tan 60° / tan 30°
FD / CF = √3 x √3
FD / CF = 3
FD = 3CF ----------------(3)
since CF + FD = 80         ( given )
=>CF + 3CF = 80
=>CF = 20 mtr    Ans.
From eq (3) 
FD = 3 x 20 = 60 mtr    Ans.
From eq (2)
AC = FD tan 30°
AC = 60 x 1/√3
AC = 60 x √3 / √3 x √3        multiply and divide by √3
AC = 20√3 mtr Ans.
11. 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° (see Fig. 9.12). Find the height of the tower and the width of the canal.
Solution:
BC = width fo the canal
AB = ? Height of the TV tower
tan 30° = AB / BD 
1/√3 = AB / (20 + BC)
20 + BC = √3AB
BC = √3AB - 20 ------------ (1)
tan 60° = AB / BC
AB = √3BC ----------------- (2) 
From eq (1)
BC = √3x√3BC - 20
BC = 3BC - 20
2BC = 20
BC = 10 mtr With of the River Ans.
From eq (2)
AB = √3BC
AB = √3x10
AB = 10√3 mtr Height of the TV tower Ans.
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:
Construction: Draw a line EB parallel to BC
Hence BE = DC
and     BC = ED = 7m (given)
Hence <ECD =<BEC = 45°     ( alternate angle)
tan<ECD = ED / DC
tan 45° = 7 / DC
1 = 7 / DC
DC = 7
tan 60° = AB / BE
 √3 = AB / BE
√3 = AB / DC        since BE = DC
√3 = AB / 7
AB = 7√3
Hence Height of the cable tower AC = AB + BC
AC = 7√3 + 7
AC = 7 (√3 + 1)m         Height of the cable tower Ans.
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:
AB = 75 m = height of the light house
CD = Distance between the ships
construction : Draw a line parallel to line DB from point A ( top of the light house )
<D = 30° = Angle of depression of second ship from top of the light house
<C = 45° = Angle of depression of first ship from top of the light house
tan 45° = AB / BC
1 = 75 / BC
BC = 75 mtr
tan 30° = AB / BD
1 / √3 = 75 / ( BC + CD )
BC + CD = 75√3
CD = 75√3 - 75
CD = 75 (√3 - 1 ) mtr Distance between both the ships. Ans.
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 Fig. 9.13). Find the distance travelled by the balloon during the interval.
Solution:
Fig 9.13



construction: draw line DB parallel to FG 
Hence, 
BD = FG
DF = BG = 1. 2 m height of the girl
AG = 88.2 mtr height of the ballon from ground
AB = EC
tan 60° = EC / CD
√3 = AB / CD    since AB = EC
√3 = ( AG - BG ) / CD
√3 = (88.2 - 1.2 ) / CD         since AG = 88.2 m and DF = BG = 1.2 m
√3 = 87 / CD
CD = 87/√3
tan 30° = AB / BD
1/ √3 = ( AG - BG ) / BD
1/√3 = 87 / BD
BD = 87√3
Distance between ballon AE = BC = BD - CD
BC = 87√3 - 87 / √3
BC = 87 ( √3 - 1/√3)
BC = 87 x 2/√3
BC = 87 x2 x √3 / √3 x √3        divide and multiple by √3
BC = 174√3 / 3
BC = 58√3  Distance between both the ballon. Ans.
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:
BC = ?
Since, Speed = Distance / Time
S=CD / 6        let speed of car is "S" m/s.
CD = 6S ----------------- (1)
tan 30° = AB / BD
1/√3 = AB / ( BC + CD )
1/√3 = AB / ( BC + 6S ) 
BC + 6S = √3AB
BC = √3AB - 6S  -------(2)
tan 60° = AB / BC
√3 = AB / BC
AB = √3BC
AB = √3(√3AB - 6S)
AB =3AB - 6√3S
2AB = 6√3S
AB = 3√3S --------------(3)
From eq (2)
BC = √3 x 3√3S - 6S
BC = 9S - 6S
BC = 3S

since speed of the car is constant i.e. "S", so the time "t" taken by the car to reach from point "B" is:
t = BC / Speed
t=3S / S
t = 3 seconds Ans.


I hope all the questions are properly readable and understandable, in case of some confusion, kindly let me know in comment section 🙏🙏🙏.




    





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