Class 10 RD SHARMA Solutions Maths Chapter 12 - Heights and Distances
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Some Applications of Trigonometry Exercise Ex. 12.1
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Let BC be the building, AB be the transmission tower, and D be the point on ground from where elevation angles are to be measured.
Solution 25
Solution 26
Let AB be the statue, BC be the pedestal and D be the point on ground from where elevation angles are to be measured.
Solution 27
Solution 28
Solution 29
Let AB be the lighthouse and the two ships be at point C and D respectively.
Solution 30
Solution 31
Solution 32
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Solution 34
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Solution 40
Solution 41
Let AC = h be the height of the chimney.
Height of the tower = DE = BC = 40 m
In ∆ABE,
∴AB = BE√3….(i)
In ∆CBE,
tan 30° = 
Substituting BE in (i),
AB = 40√3 × √3
= 120 m
Height of the chimney = AB + BC = 120 + 40 = 160 m
Yes, the height of the chimney meets the pollution control norms.
Solution 42
Let the ships be at B and C.
In D ABD,
∴ BD = 200 m
In D ADC,
Distance between the two ships = BC = BD + DC
Solution 43
Here m∠CAB = m∠FEB = 30°.
Let BC = h m, AC = x m
In D ADE,
In D BAC,
Height of the second pole is 15.34 m
Solution 44
Solution 45
Let AQ be the tower and R, S respectively be the points which are 4m, 9m away from base of tower.
As the height can not be negative, the height of the tower is 6 m.
Solution 46
Solution 47
Solution 48
Solution 49
Solution 50
Solution 51
Solution 52
Solution 53
Solution 54
Solution 55
Solution 56 (i)
Let AB be the cliff, so AB=150m.
C and D are positions of the boat.
DC is the distance covered in 2 min.
∠ACB = 60o and ∠ADB = 45o
∠ABC = 90o
In ΔABC,
tan(∠ACB)= 
In ΔABD,
tan(∠ADB)= 
So, DC=BD - BC
=
Now,
Solution 56 (ii)
Let AB be the lighthouse and C be the position of man initially.
Suppose, a man changes his position from C to D.
As per the question, we obtain the following figure
Let speed of the boat be x metres per minute.
Therefore, CD = 2x
Using trigonometry, we have
Also,
Hence, speed of the boat is 57.8 m.
Solution 57
AB is the tower.
DC is the distance between cars.
AB=120m
In ΔABC,
tan(∠ACB) =
In ΔABD,
tan(∠ADB) =
So, DC=BD+BC
Solution 58
Let CD be the tower.
So CD =15m
AB is the distance between the points.
∠CAD = 60o and ∠CBD = 45o
∠ADC = 90o
In ΔADC,
tan(∠CAD)= 
In ΔCBD,
tan(∠CBD)= 
So AB=BD - AD
Solution 59
Now, in triangle APB,
sin 60o = AB/ BP
√3/2 = h/ BP
This gives
h = 14.64 km
Solution 60
Solution 61
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Solution 78
Some Applications of Trigonometry Exercise 12.41
Solution 1
Solution 2
Solution 3
Solution 4
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Solution 6
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Solution 8
Wire BD
ED || AC
So, EA = DC and ED = AC
EA = 14
AB = EA + EB
20 = 14 + EB
EB = 6
So, the correct option is (a).
Solution 9
Solution 10
Some Applications of Trigonometry Exercise 12.42
Solution 11
Solution 12
Solution 13
If height of one person is x then height of another one is 2x. Also If angle of elevation of one is θ then for another it is 90 - θ.
AB = a
C is mid point.
So, the correct option is (d).
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
If height of one pole is x then height of the other one is 2x. Also If the angle of elevation of one is θ then for the other it is
90 - θ.
AB = a
C is mid point.
So, the correct option is (b).
Solution 19
EC || AB
Hence
EA = CB = 10
AD = AE + ED
ED = AD - AE
= 16 - 10 = 6
So, the correct option is (c).
Solution 20
Some Applications of Trigonometry Exercise 12.43
Solution 21
Solution 22
From the figure, it is cleared that we have to find the length of BC.
Solution 23
Solution 24
Solution 25
Solution 26
Solution 27
