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A tower stands at the centre of a circular park. A \: and\: B

are two points on the boundary of the park such that AB \left (=a \right ) subtends an angle of 60° at the foot of the tower, and the angle of elevation of the top of the tower from A\: or \: B \: is\: 30^{\circ} . The height of the tower is

  • Option 1)

    a/\sqrt{3}

  • Option 2)

    a\sqrt{3}

  • Option 3)

    2a/\sqrt{3}

  • Option 4)

    2a\sqrt{3}

 

Answers (1)

best_answer

As we have learned

Angle of Elevation -

If an object is above the horizontal line from the eye, we have to raise our head to view the object.

- wherein

angle of elevation

 

 

Trigonometric Ratios of Functions -

\sin \Theta = \frac{Opp}{Hyp}

\cos \Theta = \frac{Base}{Hyp}

\tan \Theta = \frac{Opp}{Base}

- wherein

Trigonometric Ratios of Functions

 

 

OAB is an equilateral triangle 

So , radius = AB = a 

Also \tan 30 \degree = \frac{h}{r} \Rightarrow h = \frac{r}{\sqrt 3 } = \frac{a}{\sqrt 3 }

 

 

 

 


Option 1)

a/\sqrt{3}

Option 2)

a\sqrt{3}

Option 3)

2a/\sqrt{3}

Option 4)

2a\sqrt{3}

Posted by

Himanshu

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