Trigonometry
Basic Knowledge
Essential Formulas:
বিবিন্ন ক োণের মোনঃ
sin𝜃 = cos(900 -𝜃)
1. sin2 + cos2 = 1
2. sin2 = 1 - cos2
3. cos2 = 1 - sin2
4. sec2 – tan2 = 1
5. sec2 = 1+ tan2
6. tan2 = sec2 -1
7. cosec2 – cot2 = 1
8. cosec2 = 1+ cot2
9. cot2 = cosec2 -1
10. tan =
=
=
11. cot =
=
12. sec =
=
13. cosec =
cos𝜃 = sin(900 -𝜃)
tan𝜃 = cot(900 -𝜃)
cot𝜃 = tan(900 -𝜃)
sec𝜃 = cosec(900 -𝜃)
cosec𝜃 = sec(900 -𝜃)
=
= tan ×cosec
=
= cot ×sec
14. sin2θ = 2 sinθcosθ =
15. cos2θ = 1 – 2sin2θ = cos2θ – sin2θ =
16. tan2θ =
17. sin3θ = 3sinθ – 4sin3θ
18. cos3θ = 4cos3θ – 3cosθ
19. sin (A+B) = sinAcos B + cosAsinB
20. sin (A-B) = sinAcos B – cosAsinB
21. cos (A+B) = cosAcos B – sinAsinB
22. cos (A-B) = cosAcos B + sinAsinB
লম্ব
Sin𝜃 = ঄বিিুজ
লম্ব
Tan𝜃 = িূবম
Sec𝜃 =
঄বিিুজ
িূবম
sin(-θ) = -sinθ
cos(-θ) = cosθ
tan(-θ) = -tanθ
cot(-θ) = -cotθ
sec(-θ) = secθ
cosec(-θ) = -cosecθ
sin (180°- θ) = sinθ
cos (180°- θ) = -cos θ
cosec (180°- θ) = cosec θ
sec (180°- θ)= -sec θ
tan (180°- θ) = -tan θ
cot (180°- θ) = -cot θ
঄বিিুজ
লম্ব
িূবম
Cos𝜃 = ঄বিিুজ
Cot𝜃 =
θ
িূবম
লম্ব
Cosec𝜃 =
঄বিিুজ
লম্ব
িূবম
Trigonometry
Degrees
Radians
0
0
30
45
600
0
=0
1
2
=
3
√ =
√ =
0
0
Sin
√0 = 0
Cos
1
Tan
0
Cot
Undefine
d
1
√3
Undefine
d
2
Sec
Cosec
0
√ =
900
বিিরে
০ কেণ ৪ ঩র্যন্ত ধোরোিোবি বলখুন
সিগুণলো মোনণ ৪ দ্বোরো িোগ রুন
4
=1
√1 = 1
সিগুণলো মোনণ িগযমল
ূ রুন
Sin এর বি঩রীিিোণি মোনগুণলো
বলখুন
0
√3
1
1
√2
2
√2
Undefine
d
0
tan =
Undefine
d
1
sec =
সূত্রোনুসোণর
tan এর বি঩রীিিোণি মোনগুণলো
বলখুন
সূত্রোনুসোণর
Sec এর বি঩রীিিোণি মোনগুণলো
বলখুন িো cosec
=
সূত্রোনুসোণর
Learning Section
1. Prove that, tanθ + cotθ = secθ.cosecθ
sin2𝜃 + cos2𝜃 = 1
Solution:
cosec𝜃 = 𝑠𝑖𝑛𝜃
tanθ + cotθ =
+
=
=
2. Proved that,
Solution:
+
=
=
+
sec𝜃 = 𝑐𝑜𝑠𝜃
= secθ.cosecθ (Proved)
×
=1
+
=
+
=
+
=
+
=
+
=
=
Exam Aid Bank Written Math (4th Edition)
+
=
=
= 1(Proved)
Page 698
Trigonometry
3. If tanθ+ sinθ = a and tanθ –sinθ = b, then prove that a2 – b2 = 4
Solution:
a2 – b2 = (tanθ+ sinθ)2 – (tanθ- sinθ)2
= 4.tanθ.sinθ
[4ab = (a+b)2 –(a-b)2]
=4
[সমোধোণনর সুবিধোণেয এ ই সোণে িগয ও িগযমূল রো িণেণে]
= 4√
[cos2 = 1 - sin2 ]
=4
= 4√
[tan =
=4
= 4√
=4
[(a2 – b2) = (a+b)(a-b)]
]
(Proved)
4. If sin4θ + sin2θ = 1 then proved that tan4θ – tan2θ = 1
Solution:
sin4θ + sin2θ = 1
Or, sin4θ = 1- sin2θ
Or, sin4θ = cos2θ
Or, tan4θ =
= sec2θ
tan4θ – tan2θ = 1 (Proved)
Or,
=
4
Or, tan θ = 1 + tan2θ
5. Prove that √
= secθ - tanθ
Solution:
√
=√
=√
=√
=
=
-
= secθ – tanθ (Proved)
6. Proved that (tanθ +secθ)2 =
Solution:
(tanθ +secθ)2 = (
+
)2
=(
)2 =
=
=
=
(Proved)
7. If cosA + sinA = √2cosA then proved that cosA –sinA = √2sinA
Solution:
cosA + sinA = √2cosA
Or, sinA = √2cosA –cosA
Or, sinA = (√2-1)cosA
Exam Aid Bank Written Math (4th Edition)
Page 699
Trigonometry
Or, (√2+1)sinA = (√2+1)(√2-1)cosA
Or,√2sinA + sinA = {(√2)2 -12}cosA
Or,√2sinA + sinA = (2-1)cosA
cosA –sinA = √2sinA
8. If tanA =
Solution:
Given that,
tanA =
then find the value of
So, cotA =
=
=
.
= √3
(
)
=
=
=
=
= ×
= (Ans.)
9. Solve: 2cos2θ + 3sinθ -3 = 0
Solution:
2cos2θ + 3sinθ -3 = 0
Or, 2(1-sin2θ) -3+ 3sinθ = 0
Or, 2(1-sinθ)( 1+sinθ) -3(1- sinθ) = 0
Or, (1-sinθ)(2+2sinθ-3) = 0
(1-sinθ)(2sinθ-1) = 0
1-sinθ = 0
Or, sinθ = 1
Or, sinθ = sin900
θ = 900
Ans: 900 Or 300
2sinθ-1 = 0
Or, sinθ =
Or, sinθ = sin300
θ = 300
10. Solve: cos2θ –sin2θ = 2 -5cosθ, when θ<900
Solution:
cos2θ –sin2θ = 2 -5cosθ
Or, cos2θ –(1-cos2θ) = 2 -5cosθ
Or, cos2θ –1+cos2θ - 2 +5cosθ = 0
Or, 2cos2θ +5cosθ -3 = 0
Or, 2cos2θ +6cosθ -cosθ -3 = 0
Or, 2cosθ(cosθ+3) -1(cosθ +3) = 0
(cosθ +3)(2cosθ -1) = 0
2cosθ -1 = 0
Or, 2cosθ = 1
Or, cosθ = = cos600
θ = 600
cosθ +3 = 0
Or, cosθ -3 [The value of cosθ cannot be less than -1]
Exam Aid Bank Written Math (4th Edition)
Ans: 600
Page 700
Trigonometry
2
11. Solve: tan θ –(1+√3)tanθ +√3 = 0
Solution:
tan2θ –(1+√3)tanθ +√3 = 0
Or, tan2θ – tanθ -√3tanθ +√3 = 0
Or, tanθ(tanθ -1)-√3(tanθ -1) = 0
(tanθ -1)(tanθ-√3) = 0
tanθ -1 = 0
Or, tanθ = 1 = tan450
θ = 450
Ans: 450 Or 600
tanθ-√3 = 0
Or, tanθ = √3 = tan600
θ = 600
12. Find A and B, if both A and B are acute angles and cot(A+ B) = 1, cot(A- B) = √3.
[PKB Officer 21]
Solution:
Given that,
cot(A+ B) = 1 ….. (i)
cot(A- B) = √3 …..(ii)
From equation(i)
cot(A+ B) = 1
Or, cot(A+ B) = cot450
From equation(ii)
cot(A- B) = √3
Or, cot(A- B) = cot300
Now, (i)+(ii)=>
A + B = 450 …. (iii)
A- B = 300 …..(iv)
2A = 750
And, (i)-(ii)=>
A=
= 37
2B = 150
B=
=7
Ans: A = 37 and B = 7
13. A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an
angle of 60° with the ground, then the height of the wall is:
Solution:
Let AB is a wall and AC is the ladder 15 m long which makes an angle of 60° with the
ground.
∴ In ΔABC, ∠B = 90°
Let the height of wall, AB = h
As per question,
sinθ =
Or, sin600 =
Exam Aid Bank Written Math (4th Edition)
Page 701
Trigonometry
Or,
=
∴ Height of the wall =
∴h=
Or, 2h = 15√3
m
m (Ans.)
14. The angle of depression of a point situated at a distance of 70m from the base of a
tower is 60°. The height of the tower isD
Solution:
Let, length of the tower AB = h meter.
Here, ∠DAC = ∠ACB = 600
Given that, BC = 70 meters.
According to ABC
tan600 =
Or, √3 =
h = 70√3
The height of the tower = 70√3 meters. (Ans.)
15. If a 1.5 m tall girl stands at a distance of 3 m from a lamp-post and casts a
shadow of length 4.5 m on the ground, then the height of the lamp-post is:
Solution:
Let AB is the girl and CD is lamp-post.
AB = 1.5 which casts her shadow EB
∴ EB = 4.5m, BD = 3m
According to AEB
tan =
tan = =
According to CDE
tan =
Or, tan =
Or, =
Or, 3h = 7.5
The height of the lamp-post = 2.5m(Ans.)
h=
= 2.5
16. The angle of elevation of the top of a tower from a certain point is 30°. If the
observed moves 20 m towards the tower, the angle of elevation of top of the tower
increases by 15°. The height of the tower is:
Solution:
Let AB be the tower and C and D be the points of observation.
Given that,
∠ACB = 30°, ∠ADB = (300+150) = 45° and CD = 20m
Let, AB = h
Exam Aid Bank Written Math (4th Edition)
Page 702
Trigonometry
According to BAC
tan450 =
Or, 1 =
According to BAD
tan300 =
AC = h
Or, =
=
Or, √3h = 20 +h
h=
Or, √3h - h = 20
=
=
=
Or, h(√3-1) = 20
= 10(√3+1) = 10(1.73+1) = 27.3 m (Ans.)
17. Two ships are sailing in the sea on the two sides of a lighthouse. The angle of
elevation of the top of the lighthouse is observed from the ships are 30° and 45°
respectively. If the lighthouse is 100 m high, the distance between the two ships is:
[PKB EO 19]
Solution:
Given that, AD = 100m, ∠B = 450 and ∠C = 300
The distance between the two ships = BD + DC = BC
In ABD,
tan∠B =
Or, tan450 =
BD = 100 m
Again, In ACD,
tan∠C =
Or, 1 =
Or, tan300 =
Or,
A
100m
[tan450 = 1]
450
300
B
C
D
==
[tan300 =
]
DC = 100 = 100×1.73 = 173
The distance between the two ships, BC = 100+173 = 273 m (Ans.)
18. D is a point on the side BC of a triangle ABC such that AD ⊥ BC, E is a point on
∠
AD for which AE: ED = 5: 1. If ∠BAD = 30° and
= 6, then ∠ACB = ?
∠
Solution:
∠ABD = 1800 – 900 – 300 = 600
[ ∠BAD = 300]
Exam Aid Bank Written Math (4th Edition)
Page 703
Trigonometry
∠
= and
∠
According to the question,
∠
=6
∠
[tanθ =
=
]
Or, × = 6
Or,
× =6
∴ BD = DC
So, AB = AC
∴ ∠ACB = 600 (Ans.)
Note: In isosceles triangle altitude divides the opposite side in two equal parts.
19. On the same side of tower, two objects are located. Observed from the top of the
tower, their angles of depression are 45° and 60°. If the height of the tower is 150 m,
the distance between the objects isSolution:
Let AB be the tower and C and D be the objects.
Given that, AB = 150 m, ∠ACB = 45° and ∠ADB = 60°
According to BAD
tan600 =
Or, √3 =
According to BAC
tan450 =
Or, 1 =
CD = 150 -
AD =
Or, AD + CD = AB
=
=
×
Or,
=
+ CD = 150
= 50(3 –1.73)= 50×1.27 = 63.5m
20. A man on the top of a vertical observation tower observes a car moving at a
uniform speed coming directly towards it. If it takes 12 minutes for the angle of
depression to change from 30° to 45°, how soon after this will the car reach the
observation tower?
Solution:
Let AB be the tower and C and D be the two positions of the car.
Here, ∠ACB = 45°, ∠ADB = 30°
Let, AB = h, CD = x and AC = y
According to ABC
tan450 =
Or, 1 =
y=h
According to BAD
tan300 =
Exam Aid Bank Written Math (4th Edition)
Page 704
Trigonometry
Or,
=
Or,
Or, x + h = √3h
Or, x + y = √3h
=
x = √3h - h = h(√3 -1)
h(√3−1) unit covered in 12 min.
1 unit covered in
min.
h unit covered in
Required time = 16
=
=
=
= 16
min
min(Ans.)
21. From the top of a tower, the angles of depression of two objects P and Q (situated
on the ground on the same side of the tower) separated at a distance of 100(3 - √3) m
are 45° and 60 ° respectively. The height of the tower isSolution:
Let, OP = a and TO = H
Given that, QP = 100(3 - √3) m
According to TOP
tan600 =
Or, √3 =
According to TOQ
tan450 =
Or, 1 =
=
H = √3a …(i)
√
Or, H = 100(3 - √3) + a
Or, √3a = 100(3 - √3) + a [From equation (i)]
Or, √3a - a = 100(3 - √3)
Or, a(√3 -1) = 100√3(√3-1)
a = 100√3
From equation (i)
H = 100√3×√3 = 300
The height of the tower = 300 m (Ans.)
22. On the same side of a tower, two objects are located. Observed from the top of the
tower, their angles of depression are 45° and 60°. If the height of the tower is 600 m,
the distance between the objects is approximately equal to:
Solution:
Let DC be the tower and A and B be the objects as shown above.
Given that,
DC = 600 m, ∠DAC = 45°, ∠DBC = 60°
According to DCB
tan600 =
√3 =
Exam Aid Bank Written Math (4th Edition)
Page 705
Trigonometry
BC =
According to DCA
tan450 =
Or, 1 =
=
AB = 600 -
Or, AB + BC = 600
=
=
=
Or, AB +
=
= 600
= 200(3-1.73)= 254m
23. Raju stands in a corner of his square farm. Angle of elevation of a scarecrow
placed in diagonally opposite corner is 60°. He starts walking backwards in a straight
line and after 80ft he realizes that angle of elevation of the scarecrow now is 30°.
What is area of the field?
Solution:
We know,
tan600 = √3
Or, = √3
PQ = √3QR
Again,
tan300 =
Or,
=
Or,
=
Or, 80 + QR = √3PQ = √3×√3QR
Or, 80 + QR = 3QR
Or, 3QR –QR = 80
QR = = 40
Here, QR is a diagonal of the square farm.
We know,
Diagonal of square = √2×one side of the square
One side of the square =
Area = (one side of the square)2 = ( )2 =
Or, 2QR = 80
= 800 sq. ft.(Ans.)
Important questions for practice
1. Prove that,
2. Prove that,
-
=0
+
=1
3. If cot4θ – cot2θ = 1 then proved that cos4θ + cos2θ = 1
4. Prove that √
= cotθ + cosecθ
Exam Aid Bank Written Math (4th Edition)
Page 706
Trigonometry
5. Prove that
= 2cosecθ
+
6. Proved that
7. The value of
8. A =
= cotA.tanB
-
?
এিং B = cot + cosec িণল, A2 – B2 এর মোন ি?
9. Solve: 2sin2θ + 3cosθ -3 = 0, when 00<θ<900
10. Solve: sinθ + cosθ = 1, when 00
11. If cos (A+B) = and cos (A-B) =
and B. [Rupali Cash (Re-exam) 18]
θ 900
, 0°<(A+B)≤90° and A>B, find the values of A
12. The angle of elevation of a ladder leaning against a wall is 60º and the foot of the
ladder is 4.6 m away from the wall. The length of the ladder is:
13. An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his
eye to the top of the tower is 30º. The height of the tower is:
14. A and B are standing on ground 50 meters apart. The angles of elevation for these
two to the top of a tree are 60° and 30°. What is height of the tree?
15. The angle of elevation of a cloud from a point 200 m above a lake is 30° and the
angle of depression of its reflection in the lake is 60°. The height of the cloud is16. A man is watching from the top of tower a boat speeding away from the tower.
The boat makes an angle of depression of 45° with the man’s eye when at a distance
of 40 meters from the tower. After 5 seconds, the angle of depression becomes 30°.
What is the approximately speed of the boat, assuming that it is running in still
water?
17. When the sun's altitude changes from 30° to 60°, the length of the shadow of a
tower decreases by 70m. What is the height of the tower?
Answer:
7. 4
8. 0
Exam Aid Bank Written Math (4th Edition)
9. 600
10. 900 Or 00
Page 707
Trigonometry
11. A = 37 and B = 7
15. 200m
12. 9.2m
16. 31.5 km/h
14. 25√3
13. 21.6 m
17. 60.55 m
Solution of practice session
1. Prove that,
Solution:
-
-
=0
=
=
==
2. Prove that,
Solution:
+
+
=
= 0 (Proved)
=1
=
+
=
+
=
+
=
= 1 (Proved)
3. If cot4θ – cot2θ = 1 then proved that cos4θ + cos2θ = 1
Solution:
cot4θ – cot2θ = 1
Or, cot4θ = 1 + cot2θ
Or,
= cosec2θ
Or,
×sin4θ =
Or, cos4θ = sin2θ
Or, cos4θ = 1- cos2θ
cos4θ + cos2θ = 1(Proved)
4. Prove that √
×sin4θ
= cotθ + cosecθ
Solution:
√
=
=√
=√
+
×
=
5. Prove that
Solution:
+
+
+
=√
=
=
+
=
+
= cotθ + cosecθ (Proved)
= 2cosecθ
=
Exam Aid Bank Written Math (4th Edition)
Page 708
Trigonometry
=
=
= 2cosecθ (Proved)
=
=
6. Proved that
Solution:
= cotA.tanB
=
=
=
×
7. The value of
Solution:
=
-
= cotA.tanB (Proved)
?
= 2[ ×
= 2[
-
×
-
]
]
[ cos600 = & sin600 =
= 2[
]
]
= 4[
]
= 4[
] = 4[
[cosAcosB –sinAsinB = cos(A+B)]
]
= 4[
] [cos(90-20)0 = sin200]
= 4 (Ans.)
8. A =
Solution:
Given that,
A=
=
এিং B = cot + cosec িণল, A2 – B2 এর মোন ি?
[cosec2 – cot2
= 1]
=
=
= cot + cosec
A2 – B2 = (cot + cosec )2 – (cot + cosec )2 = 0 (Ans.)
Exam Aid Bank Written Math (4th Edition)
Page 709
Trigonometry
2
9. Solve: 2sin θ + 3cosθ -3 = 0, when 00<θ<900
Solution:
2sin2θ + 3cosθ -3 = 0
Or, 2(1-cos2θ) + 3cosθ -3 = 0
Or, 2-2cos2θ + 3cosθ -3 = 0
Or, -2cos2θ + 3cosθ -1 = 0
Or, -(2cos2θ - 3cosθ +1) = 0
Or, 2cos2θ - 3cosθ +1 = 0 [Multiplying by (-1)]
Or, 2cos2θ - 2cosθ - cosθ +1 = 0
Or, 2cosθ(cosθ-1)-1(cosθ-1) = 0
(cosθ-1)(2cosθ -1) = 0
cosθ-1 = 0
Or, cosθ = 1 = cos00
θ = 00 [Not accepted]
Ans: 600
2cosθ -1 = 0
Or, cosθ = = cos600
θ = 600
10. Solve: sinθ + cosθ = 1, when 00 θ 900
Solution:
sinθ + cosθ = 1
Or, (sinθ)2 = (1- cosθ)2 [Square on both sides]
Or, sin2θ = 12 – 2.1.cosθ + cos2θ
Or, 1-cos2θ = 1 - 2cosθ + cos2θ
Or, -2cosθ + cos2θ + cos2θ = 1-1
Or, 2cos2θ - 2cosθ = 0
Or, 2cosθ(cosθ -1) = 0
cosθ(cosθ -1) = 0
cosθ = 0
Or, cosθ = cos900
θ = 900
Ans: 900 Or 00
cosθ -1 = 0
Or, cosθ = 1 = 00
θ = 00
11. If cos (A+B) = and cos (A-B) = , 0°<(A+B)≤90° and A>B, find the values of A
and B. [Rupali Cash (Re-exam) 18]
Solution:
Given that,
cos (A+B) =
Or, cos (A+B) = cos45o
A+B = 45o ....... (i)
cos (A-B) =
Or, cos (A-B) = cos30o
Exam Aid Bank Written Math (4th Edition)
A-B = 30o ....... (ii)
Page 710
Trigonometry
(i)+ (ii) => 2A = 75o
A=
= 37
Putting the value of A in equation (i)
37 +B = 45o
B = 450 - 37 = 7
Ans: A = 37 and B = 7 .
12. The angle of elevation of a ladder leaning against a wall is 60º and the foot of the
ladder is 4.6 m away from the wall. The length of the ladder is:
Solution:
Let AB be the wall and BC be the ladder.
Given that, ladder AC = 4.6m
As per question,
cos600 =
Or, =
∴ BC = 2×4.6 = 9.2 m
The length of the ladder = 9.2 m(Ans.)
13. An observer 1.6 m tall is 20√3 away from a tower. The angle of elevation from his
eye to the top of the tower is 30º. The height of the tower is:
Solution:
Let AB be the observer and CD be the tower.
Here, BE ⊥ CD
Then, CE = AB = 1.6m and BE = AC = 20√3m
According to DEB
tan300 =
Or, =
DE = 20
∴ CD = CE + DE = (1.6 + 20) m = 21.6 m(Ans.)
14. A and B are standing on ground 50 meters apart. The angles of elevation for these
two to the top of a tree are 60° and 30°. What is height of the tree?
Solution:
According to PQB
tan600 =
Or, √3 =
PQ = √3BQ ….(i)
According to PQA
tan300 =
Exam Aid Bank Written Math (4th Edition)
Page 711
Trigonometry
Or,
=
=
Or, 3BQ = 50 + BQ
Or, 3BQ – BQ = 50
Or, 2BQ = 50
BQ = 25
From equation (i)
PQ = 25√3
Height of the tree = 25√3(Ans.)
15. The angle of elevation of a cloud from a point 200 m above a lake is 30° and the
angle of depression of its reflection in the lake is 60°. The height of the cloud isSolution:
Let, AC = H m and OA = a m. So, RA = H +200 +200
According to CAO
tan300 =
Or, =
a = √3H
According to RCO
tan600 =
Or, √3 =
Or, √3 =
Or, 3H = H + 400
Or, 3H –H = 400
Or, 2H = 400
H = 200
Height of cloud above lake = 200+200 m = 400m (Ans.)
16. A man is watching from the top of tower a boat speeding away from the tower.
The boat makes an angle of depression of 45° with the man’s eye when at a distance
of 60 meters from the tower. After 5 seconds, the angle of depression becomes 30°.
What is the approximately speed of the boat, assuming that it is running in still
water?
Solution:
Let AB be the tower and C and D be the two positions of the boats.
Then, ∠ACB = 45°, ∠ADB = 30° and AC = 60 m
Let, AB = h
According to BAC
tan450 =
Or, 1 =
h = 60
Exam Aid Bank Written Math (4th Edition)
Page 712
Trigonometry
According to BAD
tan300 =
Or, =
=
Or, 60 + DC = √3h
Boat’s speed =
=
DC = 60√3 – 60 = 60(√3-1)
m/s= 12(1.73-1) = 12×0.73 m/s = 12×0.73×
km/h = 31.5 km/h
17. When the sun's altitude changes from 30° to 60°, the length of the shadow of a
tower decreases by 70m. What is the height of the tower?
Solution:
Let AD be the tower, BD be the initial shadow and CD be the final shadow.
Given that,
BC = 70 m, ∠ABD = 30°, ∠ACD = 60°,
Let CD = x, AD = h
According to ADC
tan600 =
Or, √3 =
According to ADB
tan300 =
h = √3x …(i)
Or, =
=
=
[From equation (i)]
Or, 3x = 70 + x
Or, 3x –x = 70
Or, 2x = 70
x = 35
From equation (i)
h = √3×35 = 35×1.73 = 60.55
The height of the tower = 60.55 m (Ans.)
Exam Aid Bank Written Math (4th Edition)
Page 713