Applications of Trigonometry Class 10 Extra Questions Maths Chapter 9 with Solutions

Extra Questions for Class 10 Maths Applications of Trigonometry with Answers

Extra Questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

You can also download NCERT Class 10 Maths Solutions to help you to revise complete syllabus and score more marks in your examinations.

Applications of Trigonometry Class 10 Extra Questions Very Short Answer Type

Question 1.
The ratio of the height of a tower and length of its shadow on the ground is √3 : 1 what is the angle of elevation of the sun ? [CBSE 2017]
Answer:
Let TW be tower and WS be its shadow.
Let ∠TSW = θ (angle of elevation)

tan θ = \(\frac{\text { TW }}{W S}=\frac{\sqrt{3}}{1}\)
⇒ tan θ = √3
⇒ θ ⇒

Question 2.
In figure, AB is a 6 m high pole and CD is a ladder inclined at an angle of 60° to the horizontal and reaches up to a point D of pole. If AD = 2.54 m, find the length of the ladder. (use √3 = 1.73) [CBSE 2016]
Answer:

AB = AD + DB = 6 m (given)
⇒ 2.54 m + DB = 6 m
⇒ DB = 3.46 m
Now, in the right ∠d ABCD.

Thus, the length of the ladder CD is 4 m.

Question 4.
A ladder, leaning against a wall, makes an angle of 60° with the horizontal. If the foot of the ladder is 2.5 m away from the wall, find the length of the ladder. [CBSE Outside Delhi 2016]
Answer:
Let AC be the ladder and the foot C is 2.5 m away from the wall AB.

cos 60° = \(\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{2.5}{h}\)
\(\frac{1}{2}=\frac{2.5}{h}\)
h = 5 m.

Question 5.
From the top of a 25 m high cliff, the angle of elevation of a tower is found to be equal to the angle of depression of the foot of the tower. Find height of the tower.
Answer:
In right ∆ABC,

⇒ y = 25 m
∴ Height of tower = x + y = 25 + 25 = 50 m.

Question 6.
An observer 1.5 m tall is 28.5 m away from a tower of height 30 m. Find the angle of elevation of the top of tower from his eye.
Answer:
In right ∆ABC,

tan θ = 1 = tan 45°
θ = 45°

Question 7.
The tops of two poles of height 16 m and 12 m are connected by a wire, the wire makes angle of 30° with the horizontal, find length of wire.
Answer:
In right ∆EDC,

Applications of Trigonometry Class 10 Extra Questions Short Answer Type-2

Question 1.
From the top of a 7 m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower. [CBSE 2017]
Answer:
Let BG be building
TW be Tower, then:

BM = x, ∠MBT = 60° ∠MBW = 45°
Draw BM ± TW
In rt. ∠d ∆BMW
tan 45° = \(\frac{W M}{B M}\) ⇒ 1 = \(\frac{7}{x}\) ⇒ x = 7 m
In rt. ∠d ∆TMB
tan 60° = \(\frac{\mathrm{TM}}{\mathrm{BM}}\) ⇒ √3 = \(\frac{h}{x}\)
⇒ h = √3x = 7√3
Height of Tower = TW = TM + MW
= (7√3 + 7)m = 7(√3 + 1)m

Question 2.
A moving boat is observed from the top of a 150 m high cliff moving away from the cliff. The angle of depression of the boat changes from 60° to 45° in 2 minutes. Find the speed of the boat in m/h. [CBSE 2017]
Answer:
Let C & D be two positions of the boat,& AB be the cliff & let speed of boat be xm/ min.

Let BC = y
∴ CD = 2x (∵ Distance = speed × time)
In ∆ABC \(\frac{150}{y}\) = tan 60°
y ⇒ \(\frac{150}{\sqrt{3}}\) = 50√3
In ∆ABD \(\frac{150}{y+2 x}\) = tan 45°
⇒ 150 = 50√3 + 2x
⇒ x = 25(3 – √3)
∴ Speed = 25(3 – √3) m/min
= 1500(3 – √3) m/hr.

Question 3.
The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If height of the tower is 50 m, find the height of the hill. [CBSE Delhi 2017]
Answer:
Let HL be Hill and TW be Tower angle of elevations ∠WLT = 30°
∠ LWH = 60°, let WL = x
In rt. ∠d ∆LWT 50
⇒ \(\frac{50}{x}\) = tan 30°
⇒ x = 50√3 ……. (i)
In rt. ∠d ∆WLH
⇒ \(\frac{h}{x}\) = tan 60°
⇒ h = 50√3 × √3 = 150 [Using (i)]
∴ Height of hill = 150 m

Question 4.
A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of hill as 30°. Find the distance of the hill from the ship and the height of the hill. [CBSE Outside Delhi 2016]
Answer:
Let AB be height of ship’s deck and CD be hill with C as top and D as base. Here angle of elevation ∠CAE = 60°.
Draw AE ⊥ CD and angle of depression ∠EAD = 30°
Let AE = BD = x
CE = y and ED = 10
Also, ∠BDA = ∠EAD = 30° [alternate angles]
In ∆ABD

Now, CD = x + y
⇒ CD = 30 + 10 = 40 m
∴ The height of the hill is 40m.

Question 5.
The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45° and 60° respectively. Find the height of the tower and the horizontal distance between the tower and the building, (use √3 = 1.73) [CBSE Delhi 2016]
Answer:
Let the height of the tower AB be h m and the horizontal distance between the tower and the building BC be x m.
So, AE = (h – 50)m

In ∆AED, tan 45° = \(\frac{\mathrm{AE}}{\mathrm{ED}}\)
⇒ 1 = \(\frac{h-50}{x}\)
⇒ x = h – 50 …………. (1)
In ∆ABC, tan 60° = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)
⇒ √3 = \(\frac{h}{x}\)
⇒ x√3 = h …….. (2)
Using (1) and (2), we get
⇒ x = √3x – 50
⇒ x(√3 – 1) = 50
⇒ x = \(\frac{50(\sqrt{3}+1)}{2}\)
= 25 × 2.73 = 68.25 m
Substituting the value of x in (1), we get
68.25 = h – 50
⇒ h = 68.25 + 50
⇒ h = 118.25 m
Hence, the height to tower is 118.25 m and the horizontal distance between the tower and the building is 68.25 m.

Question 6.
There are two poles, one each on either bank of a river, just opposite to each other. One pole is 60 m high. From the top of this pole, the angles of depression of the top and the foot of the other pole are 30° and 60° respectively. Find the width of the river and height of the other pole.
Answer:
Let AB, CD be two poles separated by river of width CA with AB = 60 m and let CD = h m

Draw DE ⊥ AB
BE = AB – EA = (60 – h) m
In right ∆BAC

Thus, width of river = 20√3 m
In right ∆BED
tan 30° = \(\frac{\mathrm{BE}}{\mathrm{DE}}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{60-h}{20 \sqrt{3}}\) ⇒ 20 = 60 – h
⇒ h = 60 – 20 = 40
Height of the other pole = 40 m.

Question 7.
From an aeroplane vertically above a straight horizontal plane, the angles of depression of two consecutive kilometres stones on the opposite sides of the aeroplane are found to be α and β. Show that the height of the aeroplane is \(\frac{\tan \alpha \tan \beta}{\tan \alpha+\tan \beta}\)
Answer:
Let A be the aeroplane and its height be h km further, let B and C be two consecutive kilometres stone so that distance BC = 1 km.
Let BD = km
Then DC = (1 – x) km

Question 8.
Two ships are there in the sea on either side of a lighthouse in such a way that the ships and the lighthouse are in the same straight line. The angles of depression of two ships as observed from the top of the lighthouse are 60° and 45°. If the height of the lighthouse is 200 m, find the distance between the two ships. [Use √3 = 1.73]
Answer:

Distance between two ships = x + y
= 115.4 + 200 = 315.4 m

Question 9.
The angles of elevation and depression of the top and the bottom of a tower from the top of a building, 60 m high, are 30° and 60° respectively. Find the difference between the heights of the building and the tower and the distance between them. [CBSE 2017]
Answer:
In right ∆ABE,

∴ Difference between heights of the building and tower = y = 20 m
Distance between tower and building = x
= 20√3 m

Applications of Trigonometry Class 10 Extra Questions Long Answer Type 1

Question 1.
The angle of elevation of a cloud from a point 60 m above the surface of the water of a lake is 30° and the angle of depression of its shadow in water of lake is 60°. Find the height of the cloud from the surface of water. [CBSE 2017]
Answer:
Let C be cloud &B be point 60 m above the surface of water angle of elevation of cloud = ∠MBC = 30°
Angle & Depression of clouds reflection
‘R’ ∠MBR = 60°
Let BM = x, CM = h, NR = 60 + h,
MR = 60 + 60 + h = 120 + h

In rt. ∠d ∆BMC
\(\frac{h}{x}\) = tan 30°
⇒ x = h√3
In rt. ∠d ∆BMR
\(\frac{60+60+h}{x}\) = tan 60°
⇒ \(\frac{120+h}{x}\) = √3
⇒ 120 + h = h√3 × √3 [using (i)]
⇒ h = 60
∵ height of cloud from surface of water = (60 + 60)m = 120 m.

Question 2.
Two points A and B are on the same side of a tower and in the same straight line with its base. The angles of depression of these points from the top of the tower are 60° and 45° respectively. If the height of the tower is 15 m, then find the distance between these points. [CBSE Delhi 2017]
Answer:
Angle of depressions are ∠XDB = 45°, ∠XDA = 60°
∠CBD = ∠XDB = 45° (alt. ∠s)
∠CAD = ∠XDA = 60° (alt. ∠s)
Let CA = x and AB = y

In rt. ∠d ∆ACD, = \(\frac{15}{x}\) = tan 60°
⇒ x = \(\frac{15}{\sqrt{3}}\) = 5√3 m ………… (i)
In rt. ∠d ∆ ABCD, \(\frac{15}{x+y}\) = tan 45°
⇒ 15 = x + y (∵ tan 45° = 1)
⇒ 15 = 5√3 + y
y = 15 – 5√3 = 5(3 – √3)
∴ Distance between points is 5(3 – √3)M.

Question 3.
An observer finds the angle of elevation of the top of the tower from a certain point on the ground as 30°. If the observer moves 20 m towards the base of the tower, the angle of elevation of the top increases by 15°, find the height of the tower. [CBSE 2017]
Answer:
Let TW be the tower and D, C be the points from where angle of elevation of the top of town is 30° and 45° respectively.

\(\frac{h}{x}\) = tan 45°
⇒ h = x …(i)
In rt. ∠d ∆TWD
⇒ \(\frac{h}{x+20}\) = tan 30°
⇒ h√3 = x + 20
⇒ h√3 = h + 20 [Using (i)]
⇒ h = \(\frac{20}{\sqrt{3}-1}\) or 10(73+1)
∴ Height of tower = 10(√3 + 1)m

Question 4.
A man is standing on the deck of a ship, which is 8 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Calculate the distance of the hill from the ship and the height of the hill.
Answer:

Let x m be the distance between hill and man. The angles of elevation and depression are 60° and 30° respectively. Various arrangements are as shown in the figure.
In right ∆DBC, \(\frac{\mathrm{AB}}{\mathrm{BD}}\) = tan 60°
\(\frac{h}{x}\) = √3
h = √3x ………….. (1)
In right ∆DBC, \(\frac{\mathrm{BC}}{\mathrm{BD}}\) = tan 60°
\(\frac{8}{x}=\frac{1}{\sqrt{3}}\)
x = 8√3 ………….. (2)
From (1) and (2), we get
h = √3 (8 √3) = 8 × 3 = 24 m
∴ Height of hill = (h + 8) m
= (24 + 8) m = 32 m
Hence, height of hill and distance of man from hill are 32 m and 8√3 m respectively.

Question 5.
As observed from the top of a 100 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. [Use √3 = 1.732] [CBSE 2018]
Answer:
Let AB be the lighthouse. C and D be the two ships C nearer and D farther from tower so that angle of depressions:

∠XAD = 30° and ∠XAC = 45°
∠ACB = ∠XAC = 45° (alt.angles)
∠ADB = ∠XAD = 30°
In ∆ABC,

⇒ 100 + x = 100√3
⇒ x = 100√3 – 100 = 100 (√3 – 1)
= 100 (1.732 – 1)
= 100 × 0.732 = 73.2 m

Question 6.
The angle of elevation of a jet plane from a point on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30° If the jet plane is flying at a constant height of 1500√3 m, find the speed of jet plane.
Answer:
Let EB = DC = 1500√3 m be the height of plane E and D are different positions of plane. Various arrangements are as shown in the figure.

Question 7.
The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 60°, respectively. Find the height of the tower and also the horizontal distance between the building and the tower. [CBSE Sample Paper 2017-18]
Answer:
In fig. Let BG be the building & TR be Tower
∠XTB = 30°, ∠XTG = 60°
∠TBP = ∠XTB = 30° [alt. angles]
∠TGR = ∠XTG = 60° [alt. angles]
In ∆BTP ⇒ tan 30° = \(\frac{\mathrm{TP}}{\mathrm{BP}}\)

Now, Tp √3 = \(\frac{T R}{\sqrt{3}}\) (as BP = GR)
⇒ 3TP = TP + PR
⇒ 2TP = BG ⇒ TP = \(\frac{50}{2}\) m = 25 m
Now, TR = TP + PR = (25 + 50) m.
Height of tower = TR = 75 m.
Distance between building and tower = GR
= \(\frac{\mathrm{TR}}{\sqrt{3}}\)
⇒ GR = \(\frac{75}{\sqrt{3}}\) m = 25 √3 m

Question 8.
The angle of elevation of a cloud from a point 60 m above a lake is 30° and the angle of depression of the reflection of cloud in the lake is 60°. Find the height of the cloud. [CBSE 2008, 11]
Answer:
Let F be the point at height of 60 m from the surface of lake.
The angle of elevation and depression are 30° and 60° respectively.
Various arrangements are as shown in figure.

According to law of reflection,
CD = AC = AB + BC = (H + 60) m
∴ BD = BC + CD = (60 + H + 60) m
= (H + 120) m
In right ∆DBF,

or 3H = H + 120
or 2H = 120
or H = \(\frac{120}{2}\) = 60 m
Now, height of cloud = AC
= (H + 60) m = (60 + 60) m = 120 m.

Question 9.
A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such a way that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°. Find the speed of flying of the bird. (Take √3 =1.732) [CBSE 2016]
Answer:
Let P and Q be the two positions of the bird, and let A be the point of observation. Let ABC be the horizontal line through A.

Given: The angles of elevations ∠PAB = 45° and
∠QAB = 30°, respectively.
∴ ∠PAB = 45° and ∠QAB = 30°
Also, PB = 80 m
In ∆ABP, we have tan 45° = \(\frac{\mathrm{BP}}{\mathrm{AB}}\)
⇒ 1 = \(\frac{80}{\mathrm{AB}}\)
⇒ AB = 80 m
In ∆ACQ, we have tan 30° = \(\frac{\mathrm{CQ}}{\mathrm{AC}}\)
⇒ \(\frac{1}{\sqrt{3}}=\frac{80}{\mathrm{AC}}\)
⇒ AC = 80 √3 m
∴ PQ = BC = AC – AB
= 80 √3 – 80 = 80(√3 – 1)
So, the bird covers 80 (√3 – 1) m in 2 s.
Thus, speed of the bird
= \(\frac{\text { Distance }}{\text { Time }}=\frac{80(\sqrt{3}-1)}{2}\) m/s
= \(\frac{40(\sqrt{3}-1) \times 60 \times 60}{1000}\) km/h
= 144 (1.732 – 1) km/h
= 105.408 km/h.

Applications of Trigonometry Class 10 Extra Questions HOTS

Question 1.
A spherical balloon of radius r subtends an angle 9 at the eye of an observer. If the angle of elevation of its centre is $, find the height of the centre of the balloon.
Answer:
Let O be the centre of the balloon of radius . OP = r. Let E be the eye of an observer so that angle subtended by balloon at the eye E is ∠PEQ = θ and angle of elevation of centre of balloon = ∠OEM = Φ
Let the height of the centre of balloon be h i.e., OM = h,
Let OE = d
From geometry ∠PEO = ∠QEO = \(\frac{\theta}{2}\)
Now, as radius is perpendicular to the tangent at point of contact .
∴ ∠OPE = 90°
So, ∆OPE is right angled at P and ∆OME is right angled at M.
In right ∆OPE,

Hence, the height of centre of the balloon is r sin Φ cosec \(\frac{\theta}{2}\).

Question 2.
At the foot of a mountain the elevation of its summit is 45°, after ascending 1000 m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. Find the height of the mountain.
Answer:

Let F be the foot and T be the summit of the mountain TFH such that ∠TFH = 45°
∴ In right ∆TBF,
∠BTF = 90° – 45° = 45°
Let height of mountain be h i.e., TB = h
since ∠TFH = ∠BTF = 45° ⇒ BF = BT = h
∠DFE = 30° and FD = 1000 m, ∠TDP = 60°
Draw DE ⊥BF and DP ⊥ BT
In right ∆DEF,
cos 30° = \(\frac{\mathrm{FE}}{\mathrm{DF}}\) ⇒ \(\frac{\sqrt{3}}{2}=\frac{\mathrm{FE}}{1000}\)
⇒ FE = 500√3 m
Also sin 3o° = \(\frac{\mathrm{DE}}{\mathrm{DF}}\) ⇒ \(\frac{1}{2}=\frac{\mathrm{DE}}{1000}\)
⇒ DE = 500 m
∴ BP = DE ⇒ BF = 500 m
Now TP = TB – BP = h – 500
In right ∆TPD,

Multiple Choice Questions

Choose the correct option out of four given in each of the following:

Question 1.
If the ratio of length of the shadow of a pole to its height is √3 : 1, then elevation of the sun is
(a) 30°
(b) 45°
(c) 60°
(d) 120°
Answer:
(c) 60°

Question 2.
A steel wire is tied to the top of an electric pole and the ground making an angle of 60° with the ground. If the height of electric pole is 12 m, then length of steel wire is
(a) 4√3 m
(b) 8√3 m
(c) \(\frac{4}{\sqrt{3}}\) m
(d) 12√3 m
Answer:
(b) 8√3 m

Question 3.
In the below figure the angle of depression is

(a) y
(b) z
(c) x
(d) both x and y.
Answer:
(c) x

Question 4.
Height of a tower is 10 m. If the sun’s altitude is 45°, then length of the shadow is :
(a) 5 m
(b) 20 m
(c) 10√2 m
(d) 10 m
Answer:
(d) 10 m

Question 5.
A bridge crossing a river makes an angle of 30° with the river bank [see fig.]. If the length along the bridge from bank to bank is 300 m, then width of river is

(a) 150 m
(b) 600 m
(c) 450 m
(d) 300√2 m
Answer:
(a) 150 m

Question 6.
A tree is broken by the wind, its top struck the ground at an angle of 30° at a distance of 30 m from its foot. The whole height of tree is
(a) 30 √3 m
(b) 20 √3 m
(c) 10√3 m
(d) 40√3 m
Answer:
(a) 30 √3 m

Question 7.
The angle of elevation of a tower from two points distant l and m (l > m) from its foot and in the same straight line from it are complementary, then the height of the tower is
(a) \(\sqrt{\frac{l}{m}}\)
(b) 2√lm
(c) √lm
(d) \(\sqrt{\frac{m}{l}}\)
Answer:
(c) √lm

Question 8.
An observer 1.5 m tall is 28.5 m away from a tower of height 30 m. The angle of elevation of the top of tower from his eye is
(a) 30°
(b) 45°
(c) 60°
(d) 75°
Answer:
(b) 45°

Question 9.
If the shadow of a tower standing on a level plane is found to be 50 m longer when sun’s elevation is 30° than when it is 60°, then the height of the tower is
(a) 25 m
(b) 50√3 m
(c) 25√m
(d) \(\frac{50}{\sqrt{3}}\)m
Answer:
(c) 25√m

Question 10.
From the top of a tower h metre high, the angle of depression of two objects, which lie on either side of it are a and p. The distance between the two objects is
(a) h (cot α + cot β)
(b) h (cot α – cot β)
(c) h (tan α + tan β)
(d) h (tan α – tan β)
Answer:
(a) h (cot α + cot β)

Fill in the blanks

Question 1.
Line joining the eye of an observer to the object viewed by observer is called line of ___________ .
Answer:
sight

Question 2.
The angle formed by the line of sight with the horizontal when it is above the horizontal level is called ___________ .
Answer:
angle of elevation

Question 3.
The angle formed by the line of sight with the horizontal when it is below the horizontal level is called ___________ .
Answer:
angle of depression

Question 4.
Trigonometry was invented because its need arose in ___________ .
Answer:
Astronomy

Question 5.
If the length of the shadow of a tree is √3 times its height, then the angle of elevation is ___________ .
Answer:
30°

Question 6.
If the angle of elevation of a tree from a certain distance on the ground is 45°, then the height of tree is equal to ___________ .
Answer:
its distance from observation point

Question 7.
The angles of elevation of the top of a tower from two points distant a and b from the base on the same straight line with it are complementary. The height of the tower is ___________ .
Answer:
√ab units

Question 8.
The angle of ___________ of sun at a given time from a point is called sun’s ___________ at that time.
Answer:
elevation, altitude

Question 9.
If the sun’s altitude increases, then the length of shadow of a tower ___________ .
Answer:
decreases

Question 10.
A pole is being broken by a storm at one of its point of trisection, then the top struck the ground at an angle of ___________ .
Answer:
30°

Extra Questions for Class 10 Maths

NCERT Solutions for Class 10 Maths

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