Question

plz solve it.
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Answer 1

SOLUTION :-

=>Let AB and CD be the poles of O is the point from where elevation angles are measured.

=>In ΔABO.

=>AB/BO = tan60°

=>AB/BO = √3

=>.°. BO = AB/√3.

=>In ΔCDO,

=>CD/DO = tan30°

=>CD/ 80-BO = 1/√3

=>CD√3 = 80 - BO

=>CD√3 = 80-AB/√3

=> .°. CD√3 + AB/√3 = 80

=>Since, the poles are of equal heights.

=> CD = AB

=>CD[√3 + 1/√3] = 80

=>CD(3+1/√3)=80

=>.°. CD = 20√3

=>BO = AB/√3 = CD/√3= (20√3/√3)m = 20m.

=> DO = BD-BO = (80-20)m = 60m.

Therefore, the height of poles is 20√3 and the point is 20m and 60m far from these poles.

Answer 2

Answer-

Given:

• The road is 80m wide
• The angle of elevation of top of the towers are 60° and 30°

Need to find:-

1. Height of towers =?
2. Distance of the point = ?

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$\huge{\underline{\underline{\red{SOLUTION-}}}}$

let the towers be AB and DE

Let c be the point and BC= x m

Then EC= (80-x)m

$\angle{DCE=30}$ and $\angle{ACB=60°}$

Now,

$\dfrac{DE}{EC} = \tan(30) \\ \implies \: \dfrac{DE}{80 - x} = \tan(30) \\ \implies \: \dfrac{DE}{80 - x} = \dfrac{1}{ \sqrt{3} } \\ \implies \: DE = \dfrac{80 - x}{ \sqrt{3} } - - - - - i$

also

$\dfrac{AB}{BC} = \tan(60) \\ \implies\dfrac{AB}{x} = \tan(60) \\ \implies \: \dfrac{AB}{x} = \sqrt{3} \\ \implies AB = \sqrt{3} \times x - - - - ii$

As the poles are of equal height,

thus i = ii

$\sqrt{3} x = \dfrac{80 - x}{ \sqrt{3} } \\ \implies \: 3x = 80 - x \\ \implies \: 4x = 80 \\ \implies \: x = 20$

$\pink{\boxed{\boxed{x=20m}}}$

Thus the distance of the point from one tower is 20m

From another tower is (80-20)m= 60m

The heights of the tower are

$\sqrt{3} x \\ \ = \sqrt{3} \times 20 \\ = 20 \sqrt{3} \\ = 34.641$

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Distance of point from one tower = 20m

Distance of the point from other tower =60m

Height of the towers=${20\sqrt{3}$m or 34.641m