1.

The angle of elevation of to be top point `P`of the vertical tower PQ of height `h`from poin A is `45^0`and from a point B, the angle of elevation is `60^0,`where B is point at a distance `d`from the point A measured along the line `A B`which makes an angle `30^0`with AQ. Prove that `d=(sqrt(3)-1)hdot`

Answer» `cosc=(a^2+b^2-c^2)/(2ab)`
`=(7^2+8^2-9^2)/(2*7*8)`
`=2/7`
`BM^2=BC^2+CM^2-2BC*CM*cosc`
`=7^2+4^2-2*7*4*2/7`
`=49`
`BM=7`
`In/_BMP`
`h/(BM)=tan15^o`
`h=BMtan15^o`
`=7tan15^o-(1)`
`tan30^o=(2tan15^o)/(1-tan^215^o`
`x=tan15^o`
`1/sqrt3=(2x)/(1-x^2)`
`1-x^2=2/3x`
`x=(-2sqrt3pmsqrt((2sqrt3)^2+4))/2`
`x=(-2sqrt3pm4)/2`
`2-sqrt3`
putting this value in equation 1
`=7(2-sqrt3)m`


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