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 🙏🙏🙏.