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Maths class 10 | Ncert maths | Height and Distance ऊंचाई और दूरी | Exercise 9.1 | Boards exam 2025
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Here are the solutions to Problems 4-7 of Exercise 9.1, Height and Distance, Class 10 Maths:

Problem 4:
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:

Let AB be the pedestal and BC be the statue.
AB = h m, BC = 1.6 m
Angle of elevation of top of statue, ∠APC = 60°
Angle of elevation of top of pedestal, ∠APB = 45°
In right ΔABP,
tan(45°) = $\frac{AB}{AP}$
AP = h m
In right ΔACP,
tan(60°) = $\frac{AC}{AP}$
$\sqrt{3} = \frac{h + 1.6}{h}$
h = $\frac{1.6}{\sqrt{3} - 1}$ m

Final Answer:
\boxed{\frac{1.6}{\sqrt{3} - 1} : m}

Problem 5:
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:

Let AB be the tower and CD be the building.
AB = 50 m, CD = h m
Angle of elevation of top of building, ∠ACB = 30°
Angle of elevation of top of tower, ∠CDA = 60°
In right ΔABC,
tan(30°) = $\frac{AB}{BC}$
BC = $50\sqrt{3}$ m
In right ΔADC,
tan(60°) = $\frac{CD}{AD}$
CD = $\frac{50}{\sqrt{3}}$ m

Final Answer:
\boxed{\frac{50}{3} : m}

Problem 6:
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:

Let AB and CD be the poles.
AB = CD = h m
BD = 80 m
Angle of elevation of top of pole AB, ∠APB = 60°
Angle of elevation of top of pole CD, ∠CPD = 30°
Let BP = x m, PD = (80 - x) m
In right ΔABP,
tan(60°) = $\frac{h}{x}$
h = $x\sqrt{3}$ m
In right ΔCDP,
tan(30°) = $\frac{h}{80-x}$
h = $\frac{80-x}{\sqrt{3}}$ m
Equating h,
$x\sqrt{3} = \frac{80-x}{\sqrt{3}}$
x = 20 m
h = $20\sqrt{3}$ m

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