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A square tower stands upon a horizontal plane. From a point in this plane, from which three of its upper corners are visible, their angular elevations are respectively 45 dergrees, 60 degrees and 45 degrees. Show that the height of the tower is to the breadth of one of its sides as √6(√5+1) to 4.

please answer fast...all help appreciated :-D

Asked by reetatyagi8 | 31st May, 2015, 12:53: PM

Expert Answer:

One face of the tower with two upper corners 'D' and 'E' is shown in the figure.

'D' makes 60^o with a point 'A' in the plane. Hence,

$tan space 60 to the power of o equals fraction numerator C D over denominator A C end fraction space i n space t r i a n g l e space A C D rightwards double arrow square root of 3 equals fraction numerator H over denominator A C end fraction rightwards double arrow A C equals fraction numerator H over denominator square root of 3 end fraction$

The corner 'E' makes an angle of 45^o with point 'A' in the plane. Hence,

$tan space 45 to the power of o equals fraction numerator E B over denominator B A end fraction space i n space t r i a n g l e space A B E rightwards double arrow 1 equals fraction numerator H over denominator A B end fraction rightwards double arrow A B equals H$

In triangle ABC, the angle ACB=135^o (using symmetry, since the angle is same 45^o with two upper corners)

Triangle ABC is redrawn below:

$U sin g space sin e space r u l e space i n space t h e space a b o v e space t r i a n g l e comma space fraction numerator sin 135 to the power of o over denominator H end fraction equals fraction numerator sin theta over denominator H divided by square root of 3 end fraction rightwards double arrow sin theta equals fraction numerator 1 over denominator square root of 6 end fraction rightwards double arrow cos theta equals fraction numerator square root of 5 over denominator square root of 6 end fraction angle A equals 180 to the power of o minus open parentheses 135 to the power of o plus theta close parentheses rightwards double arrow angle A equals 45 to the power of o minus theta sin space A equals sin open parentheses 45 to the power of o minus theta close parentheses equals fraction numerator 1 over denominator square root of 2 end fraction fraction numerator square root of 5 over denominator square root of 6 end fraction minus fraction numerator 1 over denominator square root of 2 end fraction fraction numerator 1 over denominator square root of 6 end fraction equals fraction numerator square root of 5 minus 1 over denominator 2 square root of 3 end fraction A g a i n space u sin g space sin e space r u l e space i n space t h e space t r i a n g l e comma fraction numerator sin A over denominator a end fraction equals fraction numerator sin 135 to the power of o over denominator H end fraction rightwards double arrow H over a equals fraction numerator sin 135 to the power of o over denominator sin A end fraction rightwards double arrow H over a equals fraction numerator 1 open parentheses 2 square root of 3 close parentheses over denominator square root of 2 open parentheses square root of 5 minus 1 close parentheses end fraction equals fraction numerator square root of 6 over denominator square root of 5 minus 1 end fraction equals fraction numerator square root of 6 open parentheses square root of 5 plus 1 close parentheses over denominator open parentheses square root of 5 minus 1 close parentheses open parentheses square root of 5 plus 1 close parentheses end fraction equals fraction numerator square root of 6 open parentheses square root of 5 plus 1 close parentheses over denominator 4 end fraction$

Answered by satyajit samal | 1st Jun, 2015, 11:59: AM

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