document.write( "Question 1014220: There are two towers. One at A and one at B. A and B are are two points on the same horizontal ground and 120 m apart. The angle of elevation of the top of the tower at B as observed from A is twice that of the tower at A as observed at B. At C, midway between A and B, the angles of elevation are complimentary. Find the heights of the the towers.\r
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Algebra.Com's Answer #630622 by ikleyn(52803)\"\" \"About 
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\n" ); document.write( "There are two towers. One at A and one at B. A and B are are two points on the same horizontal ground and 120 m apart.
\n" ); document.write( "The angle of elevation of the top of the tower at B as observed from A is twice that of the tower at A as observed at B.
\n" ); document.write( "At C, midway between A and B, the angles of elevation are complimentary. Find the heights of the the towers.\r
\n" ); document.write( "\n" ); document.write( "I really need the answers ASAP. I'm going to pass this take home quiz tomorrow at 8:30 AM, Philippine time (GMT +8).
\n" ); document.write( "I suck at math, so I don't know what to do. Thank you so so so much :)
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\n" ); document.write( "\n" ); document.write( "Answer. The lower tower is 40 m height. The higher tower is 90 m height.\r
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\r\n" ); document.write( "The scheme is shown in the Figure to the right.\r\n" ); document.write( "The vertical lines AE and BF represent the towers.\r\n" ); document.write( "The angle ECF formed by green lines is the right angle.                        \r\n" ); document.write( "\r\n" ); document.write( "\r\n" ); document.write( "Due to this reason, the triangles ACE and BFC are similar \r\n" ); document.write( " (and their respective vertices are listed in an corresponding order).\r\n" ); document.write( "\r\n" ); document.write( "From the triangles similarity, we have a proportion \"abs%28AE%29%2Fabs%28AC%29\" = \"abs%28BC%29%2Fabs%28BF%29\", or \r\n" ); document.write( "\"h%2F60\" = \"60%2FH\". (1)\r\n" ); document.write( "\r\n" ); document.write( "In addition to it, from the condition we have these equalities:\r\n" ); document.write( "\r\n" ); document.write( "h = \"120%2Atan%28beta%29\" and H = \"120%2Atan%28alpha%29\" = \"120%2Atan%282%2Abeta%29\". (2)\r\n" ); document.write( " \r\n" ); document.write( "\r\n" ); document.write( "
\r\n" ); document.write( "\r\n" ); document.write( "Substituting (2) into (1) we get this equation: \"%28120%2Atan%28beta%29%29%2F60\" = \"60%2F%28120%2Atan%282%2Abeta%29%29\". \r\n" ); document.write( "\r\n" ); document.write( "After canceling by 60, you will get \"2%2Atan%28beta%29\" = \"1%2F%282%2Atan%282%2Abeta%29%29\", or \"4%2Atan%28beta%29\" = \"1%2Ftan%282%2Abeta%29\". (3)\r\n" ); document.write( "\r\n" ); document.write( "Replacing here \"tan%282%2Abeta%29\" = \"2%2Atan%28beta%29%2F%281-tan%5E2%28beta%29%29\", you will get the equation\r\n" ); document.write( "\r\n" ); document.write( "\"4%2Atan%28beta%29\" = \"%281-tan%5E2%28beta%29%29%2F%282%2Atan%28beta%29%29\". (4)\r\n" ); document.write( "\r\n" ); document.write( "To solve it, introduce new variable x = \"tan%28beta%29\". You will get \r\n" ); document.write( "\r\n" ); document.write( "4x = \"%281-x%5E2%29%2F%282x%29\" ---> \"8x%5E2\" = \"1+-+x%5E2\" ---> \"9x%5E2\" = \"1\" ---> x = \"1%2F3\".\r\n" ); document.write( "\r\n" ); document.write( "Thus \"tan%28beta%29\" = \"1%2F3\", and you just have h = \"120%2Atan%28beta%29\" = \"120%2A%281%2F3%29\" = 40 m.\r\n" ); document.write( "\r\n" ); document.write( "Next, \"tan%282%2Abeta%29\" = \"2%2Atan%28beta%29%2F%281-tan%5E2%28beta%29%29\" = . . . = \"3%2F4\", and you finally get \r\n" ); document.write( "\r\n" ); document.write( "H = \"120%2Atan%282%2Abeta%29\" = \"120%2A%283%2F4%29\" = 90 m.\r\n" ); document.write( "

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