Line Of Sight: An Imaginary line drawn between the object and the eyes of the observer.
The angle of elevation of the point viewed by the observer is made by the line of sight with above the horizontal level.
note: In this situation, we always raise our heads to watch the object
Angle Of Depression: The angle of depression of an object is the angle made by the line of sight and below the horizontal level.
note: In this situation, We always down our heads to look at the object.
Exercise 9.1
Q1. A circus artist climbing a 20m 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 30º
Given:
In Δ ABC
∠B = 90º
θ = 30º
AC = 20m
To Find:
Height of the pole AB =?
Solution:
AB/AC = P/H
=> AB/20 = sinθ
=> AB = 20sin30º
=> AB = 20×1/2 { sin30º=1/2 }
=> AB = 10m
Hence Height of the pole = 10meters
Q2. 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 30º with it. The distance between the foot of the tree to the point where the top touches the ground is 8m. Find the height of the tree.
Given:
According to figure
In Triangle ABC
angle of elevation θ = 30º
BC = 8m
To Find:
Height of the tree (AB+AC) =?
Solution:
In triangle ABC
∠B = 90º
BC = 8m
AC = ?
AB =?
AC/BC = hypotenuse(H)/Base(b)
=> AC/8 = secθ
=> AC/8 = sec 30º
=> AC = 8 sec 30º
=> AC = 8 × 2/√3
=> AC = 16/√3 ………………equation (i)
now for AB = ?
AB/BC =perpendicular(p)/Base(b)
=> AB /8 = tanθ
=> AB/8 = tan 30º
=> AB = 8 × 1/√3
=> AB = 8/√3 ………………………..equation (ii)
By Adding Equation(i)and Equation(ii)
AB+ AC = 8/√3 + 16/√3
=> AB + AC = 24/√3
=> AB + AC = 24/√3 ×√3 /√3
=> AB + AC = 24√3/3
=> AB + AC = 8√3
Hence Height of tree = 8√3 m
Q3. A contractor plants to instant two slides for the children to play in a park. For children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5m and is inclined at an angle of 30º to the ground, whereas for elder children she wants to have a steep slide at an angle if 60º to the ground what should be the length of the slide in each case?
Given:
Case Ist: Below the age of 5 years children
In ΔABC
AB=1.5m
θ=30º
Ac=?
In Case IInd: For 5 years elder children
In ΔPQR
PQ= 3m
θ=60º
PR=?
Solution:
For case Ist:
AC/AB = H/P
=>AC/1.5 = Cosecθ
=>AC/1.5 =cosec30º
=>AC = 1.5×2
=>AC= 3.0
=>AC=3m
For Case IInd:
PR/PQ = H/P
=>PR/3 = cosecθ
=>PR/3 = cosec60º
=>PR/3 = 2/√3
=>PR = 6/√3 ×√3 /√3
=>PR=6√3 /(√3 )² {After rationalising the denominator}
=>PR = 6√3 /3
=>PR=2√3 m
Hence
- Length of the slide for the age of 5 years = 3 meters
- Length of the slide for the elder age of 5 years = 2√3 meters
Q4. The angle of elevation of the top of a tower From a point om the ground, which is 30m away from the foot of the tower, is 30°. Find the height of the tower.
Given:
In ΔABC
BC=30m
θ=30º
To Find:
Height of the tower AB = ?
Solution:
In ΔABC
AB/BC = P/b
=>AB/30 = tanθ
=> AB/30 = tan30º
=>AB/30 = 1/√3
=>AB = 30×1/√3
=>AB=30×1/√3 × √3 /√3 {After Rationalise the denominator }
=>AB=30√3 /(√3 )²
=>AB=30√3 /3
=>AB=10√3 m
Hence height of the tower 10√3 meters
Q5. A kite is Flying at a height of 60m above the ground. The string attached to the kite is temporarily tied to a point on the ground is 60º. Find the length of the string. Assuming that there is no slack in the string.
Given:
In ΔABC
θ=60°
AB=60m
To Find:
The length of the string AC=?
Solution:
In ΔABC
AC/AB=H/P
=>AC/60=Cosecθ
=>AC/60=Cosec60°
=>AC=60× 2/√3
=>AC=120/√3 × √3/√3 { Rationalise the denominator }
=>AC=120√3/(√3)²
=>AC=120√3/3
=>AC=40√3
Hence the length of the string AC= 40√3 meters
Q6. A 1.5m tall boy is standing at some distance From a 30m tall building. The angle of elevation From his eyes to the top of the building increases From 30º to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution:
According to the figure
EF=BC=1.5m
Height of building(AC)=30m
AB=AC-BC
=>AB=30-1.5
=>AB=28.5m
In ΔABG
∠B=90°
∠G=θ=60°
AB=28.5m
BG=?
BG/AB = b/P
=>BG/AB = Cotθ
=>BG/28.5 = Cot60°
=>BG/28.5 = 1/√3
=>BG=28.5/√3 × √3/√3 { Rationalize the denominator }
=>BG=28.5√3/(√3)²
=>BG=28.5√3/3
=>BG=9.5√3 ………………….(i)
In ΔABF
∠B = 90°
∠F = θ = 30°
BF =?
BF/AB = b/P
=>BF/AB = Cotθ
=> BF/28.5 = Cot 30°
=>BF /28.5 = √3
=> BF = 28.5√3 ………………….(ii)
FG= BF – BG
=> FG = 28.5√3 – 9.5√3
=>FG = 19.0√3
=>FG = 19√3
Hence The distance he walked towards the building = 19√3 meters
Q7.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 20m high building are 45° and 60º respectively. Find the height of the tower.
Given:
Height of building BC=20m
To Find:
Height of tower=?
Solution:
According to figure
Let the Height of the tower be = x meters
Now in ΔBCD
∠C = 90º
∠BDC=θ =45º
CD =?
CD/BC = b/P = Cotθ
=> CD/ 20 = Cot45º
=> CD/20 = 1
=>CD = 20 ……………………(i)
In ΔACD
AC=AB+BC
=>AC= x+20
θ=60º
DC=?
DC/AC = b/P
=>DC/x+20 = Cotθ
=>DC/x+20 = Cot60°
=>DC = x+20 ×1/√3
=>DC = x+20 / √3 …………..(ii)
Equation(i) = Equation(ii)
CD=DC
=>20=x+20/√3
=>20√3 = x+20
=>20√3-20 = x
=>20(√3-1) =x
Hence Height of the tower = 20(√3-1)meters
Q8. A Statue,1.6m 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 60º and From the same point, the angle of elevation of the top of the pedestal is 45º. Find the height of the pedestal.
Given:
According to figure
Height of statue AB=1.6m
∠ADC = 60º
∠BDC = 45º
To Find:
Height of Pedestal BC= x (Let)
Solution:
In ΔACD
AC=AB+BC
=>AC= 1.6+X
θ=60º
DC=?
DC/AC=b/p
=>DC/1.6+x = Cotθ
=>DC/1.6+x = Cot60°
=>DC/1.6+x = 1/√3
=> DC = 1.6+x /√3 ……………………….(i)
In ΔBCD
BC=x
θ=45º
DC=?
DC/BC = b/P
=>DC/x = Cotθ
=>DC/x=Cot45º
=>DC/x = 1
=>DC =x …………………………(ii)
Equation (i) = Equation(ii)
DC=DC
=>1.6+x /√3 = x
=>1.6+x =√3 x
=> 1.6 = √3 x – x
=>1.6 = (√3 – 1)x
=>1.6/ (√3 – 1) = x
=>1.6/ (√3 – 1) × (√3 + 1)/ (√3 + 1) = x Rationalize the denominator }
=>1.6 (√3 + 1))/(3-1)= x
=>1.6 (√3 + 1)/2 =x
=>0.8 (√3 + 1)=x
Hence Height of the Pedestal = 0.8( (√3 + 1)meters
Q9. The angle of elevation of the top of a building From the foot of the tower is 30º and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50m high, Find the height of the building.
Given:
According to figure
Height of tower CD = 50m.
∠ACB = 30º
∠CBD = 60º
To Find:
Height of the building (AB)=?
Solution:
AB=Let the Height of the building be = x meters
In ΔDCB
θ=60°
DC = 50 m
BC= ?
BC/DC =b/P
=> BC/50 = Cotθ
=>BC/50 = Cot60º
=>BC/50 = 1/√3
=>BC = 50/√3 ……………………(i)
In ΔABC
AB= x (Let)
θ= 30°
BC=?
BC/AB = b/P
=>BC/x = Cotθ
=>BC/x = Cot30º
=>BC/x = √3
=>BC= √3 x …………………(ii)
Equation(i)=Equation(ii)
50/√3 = √3x
=>50/√3 = √3x
=> 50 =(√3)²x
=>50 = 3x
=> 50/3 = x
=> 16.66m
Hence the Height of the building is 16.666m
Q10. Two poles of equal height are standing opposite each other on either side of the road, which is 80m wide. From a point between them on the road, the angles of elevation of the top of the pole are 60º and 30° respectively. Find the height of the poles and the distances of the point From the poles.
Given:
According to figure
AB and CD are poles
BD= Distance between two poles =80m
∠AEB= 60º
∠CED = 30º
To Find:
(i)Height of poles = AB = CD = ?
(ii)Distances of the points E From the poles=?
Solution:
In ΔCDE
ED = x (Let)
θ = 30º
CD =?
CD/ED = P/b
=> CD/ED = tanθ
=> CD/x = tan 30º
=> CD = x tan 30º
=> CD = x × 1/√3
=> CD = x/√3 ……..(i)
Similarly In Δ AEB
θ = 60º
BE = (80 – x)meter
AB =?
AB/BE = P/b
=> AB/(80-x) = tan θ
=> AB/80 – x = tan 60º
=> AB = (80 – x)√3 …..(ii)
We know that Height of both poles are equal
Equation (i) = Equation (ii)
CD = AB
=> x/√3 =(80 – x) √3
=> x = (80 – x)√3×√3
=> x = (80 – x)3
=> x = 240 – 3x
=> x + 3x = 240
=> 4x = 240
=> x = 240/4
=> x = 60
=> x = 60 put into Equation (ii)
AB = (80 – 60) √3
AB = 20 √3
Hence:
(i) Height of each pole = 20√3
(ii) Distances of point E From the poles are 60meters and 20 meters.
Q11. 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 60º. From another point, 20m away From this point o the line joining the point to the foot of the tower, the angle of the top of the tower is 30º(see in figure) Find the height of the tower and the width of the canal.
Solution:
According to figure:
AB = TV tower
CB = width of the canal.
∠ACB = 60º
∠ADB = 30º
To Find:
(i) Height of the tower (AB) =?
(ii) Width of the canal CB =?
Solution:
In Δ ABC
θ = 60º
BC = x (Let )
AB = ?
AB/BC = P/b
=> AB/x = tan θ
=> AB/x = tan 60º
=> AB = √3 x …..(i)
In Δ ABD
BD = BC + CD
=>BD= (x + 20)
θ = 30º
AB =?
AB/BD = p/b
=> AB/BD = tan θ
=> AB/x+20 = tan 30º
=> AB = (x+20)×1/√3
=> AB = (x+20)/√3 …..(i)
Now Equation (i) = Equation(ii)
AB = AB
=> √3x = (x+20)/√3
=> 3x = (x+20)
=> 3x – x = 20
=> 2x =20
=> x = 20/2
=> x = 10 Put into equation (i)
AB = √3×10
=> AB = 10√3 meters
Hence: (i) Height of tower AB = 10√3m
(ii)Width of the canal CB = 10m
Q12. From the top of a 7m high building, the angle of elevation of the top of a cable tower is 60º and the angle of depression of its foot is 45°. Determine the height of the tower.
Given:
According to figure
Height of the building AB = 7m
∠CAE = 60º
CD = cable tower
∠EAD = 45º
To Find:
Height of the tower CD =?
Solution:
In Δ AED
∠AED = 90º
∠EAD = θ = 45º
Here AB = ED = 7m
AE = BD =?
AE/ED = b/p
=> AE/7 = cot θ
=> AE/7 = cot 45º
=> AE = 7 × 1
=> AE = 7
Now In Δ CEA
AE = 7
θ = 60º
CE = ?
CE/AE = P/b
=> CE/7 = tan θ
=> CE/7 = tan 60º
=> CE/7 = √3
=> CE = 7√3
Hence:
Height of tower CD = CE + ED
=> Height of tower CD = 7√3 + 7
=> Height of tower CD = 7(√3 + 1)m
Q13. As observed from the top of a 75 m lighthouse from the sea level, the angles of depression of two ships are 30º and 45º. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Given:
According to figure
AB = Height of lighthouse = 75 m
∠EAD = 30º
∠EAC = 45º
To Find:
Distance between two ships = CD =?
Solution:
In Δ ABC
AB = 75 m
∠EAC = ∠ACB = θ = 45º {Alternate interior angle}
BC =?
BC/AB = b/p
=> BC/AB = cotθ
=> BC/75 = cot 45º
=> BC = 75 × 1
=> BC = 75 m ……….(i)
In Δ ABD
AB = 75 m
∠EAD = ∠ADB = θ = 30° {Alternate interior angle}
BD =?
BD/75 = cot 30º
=> BD = 75√3 ………..(ii)
CD = BD – BC
=> CD = 75√3 – 75
=> CD = 75(√3 – 1)
Hence Distance between two ships = 75(√3 – 1) m
Q. 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 balloon’s angle of elevation from the girl’s eyes at any instant is 60º. After some time, the angle of elevation reduces to 30º. Find the distance travelled by the balloon during the interval.
Given:
According to figure
AC = 88.2 m
ED = BC = 1.2 m
To Find:
Distance Travelled by the balloon FB =?
Solution:
According to figure
In Δ ABC
AB = AC – BC
=> AB = 88.2 – 1.2
=> AB = 87 m
θ = 30º
BE =?
BE/AB = b/p
=> BE/87 = cot θ
=> BE/87 = cot 30º
=> BE = 87 × √3
=> BE = 87√3 ………(i)
In Δ GFE
θ = 60º
GE = AB = 87
FE =?
FE/AB = b/P
=> FE/AB = cot θ
=> FE/AB = cot 60º
=> FE/87 = 1/√3
=> FE = 87 × 1/√3
=> FE = 87/√3 ………(ii)
FB = EB – EF
=> FB = 87√3 – 87/√3 { From equation (i) and equation (ii) }
=> FB = (87×3 – 87) /√3 { taking √3 as LCM }
=> FB = (261 – 87)√3
=> FB = 174/√3 × √3/√3
=> FB = 174√3/3
=> FB = 58√3 m
Hence the Distance travelled by the balloon = 58√3 m.
Q.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 60º. Find the time taken by the car to reach the foot of the tower from this point.
Given:
According to figure
DC = tower
∠EDA = ∠DAC = 30º
∠EDB = ∠DBC = 60º
To Find :
Time is taken by car to reach the foot (c) of the tower From point B =?
Solution:
According to figure:
Let the uniform speed of the car be = x m/sec.
AB = Distance covered by car in 6 seconds =speed × time
=> AB = Distance covered by car in 6 seconds = x × 6=6x m
Let time taken by car from B to C be = t seconds
∴ Distance BC = speed × time =
=>Distance BC = x × t =x t
In triangle BCD
∠DBC = ∠EDB =θ =60º
DC =?
DC/BC = perpendicular(p)/base(b)
=> DC/x t = tanθ
=>DC/x t = tan60º
=> DC = √3×xt
In Δ ACD
AC = AB + BC =6x +xt
θ =30º
DC =?
DC/AC = Perpendicular(p)/Base(b)
=> DC/AC = tanθ
=>DC/6x+xt = tan30º
=> DC/6x+xt =1/√3
=> DC = (6x+xt)×1/√3
=> DC = (6x+xt)/√3 ……………..equation(ii)
Equation(i) = Equation(ii)
√3 xt = (6x+xt)/√3
=> √3 x t ×√3 = (6x+xt)
=> 3 x t = 6x + x t
=> 3xt – x t = 6x
=> 2xt = 6x
=> t =6x/2x
=> t =3 seconds
Hence time taken by car to reach the foot of tower from point B = 3seconds
Q16. The angle of elevation of the top of a tower From two points at a distance of 4m and 9m.From the base of the tower and in the same straight line with it are complementary .Prove that the height of the tower is 6m.
Given:
According to figure
AB = Height of tower
BC = 4m
BD = 9m
∠ACB + ∠ADB = 90º
To Prove:
Height of tower AB = 6m
Solution:
In triangle ABC
BC = 4m
∠ACB = θ
AB = ?
AB/BC = perpendicular(p)/base(b)
=> AB/4 = tanθ
=> AB = 4 tanθ …………………………….equation(i)
In triangle ABD
∠ADB = 90º – θ
BD = 9m
AB =?
AB/BD = perpendicular(p)/base(b)
AB/9 = tan(90º-θ)
=> AB = 9 cotθ {by Applying tan(90º-θ) = cotθ } ……………………..equation(ii)
equation(i) × equation(ii)
AB×AB = 4 tanθ × 9 cotθ
=> (AB)² = 4×9 × tanθ × cotθ
=> (AB)² = 4×9 × tanθ × 1/tanθ { by applying cotθ = 1/tanθ }
=> (AB)² = 36
=> AB = √36
=> AB = 6
Hence Height of tower = 6m
Hence proved