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`5 sqrt(2/3)``(5sqrt(3))/2``5sqrt(3/2)``5sqrt(6)`

Solution :

Let AB be the pole standing on the building BC and O be the position of the antenna .<br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CEN_TRI_C06_E01_013_S01.png" width="80%"> <br> In triangle OAB ,<br> `tanalpha =5/x` …(1) <br> In triangle OAC ,<br> `tan2alpha = (30)/x` ...(2) <br> From (1) and (2) . <br> `tan2alpha = 6tanalpha ` <br> `rArr 3-3tan^2alpha =1` <br> `rArr tanalpha = sqrt((2)/(3))` <br> From (1) , `x= 5cotalpha =5sqrt((3)/(2))`Transcript

Time | Transcript |
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00:00 - 00:59 | hello students A5 M Hai Pol stands on a building of height 25 m the poet and the building subtend equal angles at an antenna placed at a height of 30 M the distance of the antenna from the top of the Pole is what sunao please understand that the angle that is being subtended is taken from the base of the this case would be and the base of the building which in this case is C ok and we have to find X are right now in triangle abc we can see that tan Alpha is equal to 5 by X ok this implies X is equals to 5 Court Alpha it is called this equation 1 ok similarly in triangle AOC turn off Two Alpha |

01:00 - 01:59 | request to 30 by X ok now latest call this equation to from one and two we can turn off Two Alpha is equals to 6 x 10 Alpha right now what do you want to do is begin to solve for tan alpha and we get 3 - 3 tan Alpha square equal to 1 which implies tan Alpha is equal to under root 2 by 3 now equation one we know that it is equal to 5 Quad Alpha this implies X is equals to 5 root 3 by 2 which medicine is this one |

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