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`        The angle of elevation of the top point P of the vertical tower PQ of height ‘h’ from a point A is 45° and from a point B, the angle of elevation is 60°, where B is a point at a distance ‘d’ from the point A measured along the lines AB which makes an angle 30° with AQ. Prove that d =h(3^1/2 -1)`
10 months ago

```							I cannot attach an image.But follow the steps one by one.Draw a triangle APQ.angle PQA=90.PQ=h.angle PAQ=45.Say AQ=l.h/l=tan45Or,l=hNow draw a point inside the triangle.This point is B.join AB.AB=d.Drop a perpendicular from B on AQ.name the touching point M.This will from another triangle ABM.angle BAM=30.say AM=x and BM=y.say BN=x'.SO l=x'+xOr,h=x'+xWe can find out x from triamgle ABM.cos30=x/dOr,x=root(3)d/2Sin30=y/dOr,y=d/2Join PB.drop a perpendicular from B on PQ.name the touching point N.angle PBN=60.NQ=y.PN=h-y.tan60=(h-y)/x'x'=(h-d/2)/root3Put these values in l=h=x+x' and you will get answer.
```
10 months ago
```							Typing here is really painful.I typed at one place,it appeared at another place.After "BM=y" please do the drawing "join PB".then go back to BN=x' and  do each step.
```
10 months ago
```							In ΔAPQ, Q is right angle, PAQ=π/4, PQ=h (given)PQ/AQ=tan(π/4)=1 => AQ=PQ=hLocate point B in this Δ such that BAQ= π/6From B, drop perpendiculars onto PQ & AQ respectively cutting them at R & C.Thus ABC & PBR are two Δs with AB=d & BCQR is a rectangle. In triangle ΔABC, AC=d*cos(π/6) = d√3/2, BC = d*sin(π/6)=d/2Also, BR=CQ=AQ-AC=(h-d√3/2)=(2h-d√3)/2QR=BC=d/2; PR=PQ-QR=(h-d/2)=(2h-d)/2PBR = π/3 (given) In ΔPBR, PR/BR=tan(π/3)=> [(2h-d)/2]/[(2h- d√3)/2]= √3=> (2h-d)/(2h- d√3)= √3=> 2h-d= √3*(2h- d√3)=> 2h-d = (2√3h-3d)=> 2h*(√3-1)=3d-d=2d=> d= h*(√3-1) Trust this helps.
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one month ago
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