Question

The angle of elevation of the top of a lower 30 m high from the foot of another tower in
the same plane is 60°, and the angle of elevation of the top of the second tower from the
foot of the first tower is 30°. Find the Distance between the two towers and also the
height of the other tower.
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Answer 1

Answer:

Chapter wise Important Questions for CBSE Class 10 Maths. Chapter 1 - Real Numbers. Chapter 2 - Polynomials. Chapter 3 - Pair of Linear Equations in Two Variables. Chapter 4 - Quadratic Equations. Chapter 5 - Arithmetic Progressions. Chapter 6 - Triangles. Chapter 7 - Coordinate Geometry.

Real Numbers

Answer 2

★ Question :-

• The angle of elevation of the top of a lower 30 m high from the foot of another tower in the same plane is 60°, and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the Distance between the two towers and also the height of the other tower.

★ Answer

Given :-

• The angle of elevation of the top of a lower 30 m high from the foot of another tower in the same plane is 60°.
• The angle of elevation of the top of the second tower from the foot of the first tower is 30°

To Find :-

• Distance between the two towers
• The height of the other tower.

Formula used :-

$\bf \:tanx = \dfrac{perpendicular}{base}$

Solution :-

Let 'h' meter be the height of the other tower and distance between two towers be 'x' m.

★In right triangle AEB

$\bf\implies \:tan60 = \dfrac{AE}{AB}$

$\bf\implies \: \sqrt{3} = \dfrac{30}{x}$

$\bf\implies \:x = \dfrac{30}{ \sqrt{3} }$

$\bf\implies \:x = \dfrac{30}{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} }$

$\bf\implies \:x = 10 \sqrt{3} \: m$

★ This implies, distance between two towers is 10 √3 m.

★ Now, in right triangle ABC

$\bf\implies \:tan30 = \dfrac{BC}{AB}$

$\bf\implies \:\dfrac{1}{ \sqrt{3} } = \dfrac{h}{10 \sqrt{3} }$

$\bf\implies \:h = 10 \: m$

★ This, implies the height of other tower is 10 m.