Question

Q5. The angle of elevation of the top of a tower from a point 30 m away from its base is 30°. The height of the tower is
1)10/√3m
2)20√3m
3)30√3m
4)10√3m​

Answer 1

Answer:

this is your correct answer

Answer 2

Answer :

Diagram -

$\setlength{\unitlength}{20} \begin{picture}(5,5) \linethickness{1}\put(1,1){\line(1,0){3}} \put(4,1){\line(0,1){4}}\qbezier(1,1)(1,1)(4,5)\qbezier(1.7,1)(1.7,1.4)(1.5,1.6)\put(3.7,1){\line(0,1){0.3}}\put(3.7,1.3){\line(1,0){0.3}}\put(3.9,5.1){ \bf A }\put(3.9,0.5){ \bf B }\put(0.8,0.5){ \bf C }\put(1.8,1.25){ \bf {30}^{ \circ} }\put(2,0.5){ \bf 30 \: m }\put(4.2,3){ \bf ? }\end{picture}$

Let tower be AB and point be C.

Given -

• The distance of point C from the tower, BC = 30 m.

• Angle of elevation, $\angle$ ACB = 30°

As we can see that the tower is vertical. Therefore, $\angle$ ACB = 90°.

To find-

• The height of the tower

Solution -

In right - triangle ABC,

$\\ \sf \: \tan(c) = \dfrac{side \: \: opposite \: to \: \angle \: c}{side \: \: adjacent \: \: to \: \angle \: c} \\ \\ \\ \implies \sf \: tan \: {30 {}^{ \circ}} = \dfrac{AB}{BC} \\ \\ \\ \implies \sf \: \dfrac{1}{ \sqrt{3} } = \dfrac{AB}{30} \\ \\ \\ \implies \sf \: AB = \dfrac{30}{ \sqrt{3} }$

Now, by multiplying √3 in both numerator and denominator -

$\\ \sf \: AB = \dfrac{30}{ \sqrt{3} } \times \dfrac{ \sqrt{3} }{ \sqrt{3} } \\ \\ \\ \implies \sf \: AB = \frac{ \not{30} \sqrt{3} }{ \not 3} \\ \\ \\ \implies \large{ \boxed{ \sf \: AB = 10 \sqrt{3} }}$

Therefore, right option is a.) 10√3