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Potwe 13 Gm Fc 06 s

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  WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Problem of the WeekProblem E and SolutionsIn the Spotlight Problem In an art gallery, a 2 m high painting,  BT  , ismounted on a wall with its bottom edge 1 m abovethe floor. A spotlight is mounted at  S  , 3 m outfrom the wall and 4 m above the floor. Determine ∠ TSB , accurate to 1 decimal place. 2 m1 m3 m4 m T W  F S  B Solution 1 Let  X   be the point on  SF   such that  BX   isparallel to  WF  . Then  BX   =  WF   = 3, XF   =  WB  = 1, and  XS   = 3 follows. ∴   BXS   is an isosceles right-angled triangle, andso  ∠ BSX   = 45 ◦ . 213331 T W  F S  B X  Let  Y    be the point on  SF   such that  TY    is parallelto  WF  . Then  TY    =  WF   = 3,  YF   =  WT   = 3,and  YS   = 1 follows. Since   TYS   is right-angled,tan( ∠ TSY   ) =  TY  SY    =  31  = 3 and so  ∠ TSY    ≈ 71 . 6 ◦ . ∴ ∠ TSB  =  ∠ TSY    − ∠ BSX   ≈ 71 . 6 ◦ − 45 ◦  ˙= 26 . 6 ◦ .  33331 Y T W  F S   WWW.CEMC.UWATERLOO.CA | The CENTRE for EDUCATION in MATHEMATICS and COMPUTING Solution 2 Let  X   be the point on  SF   such that  BX   isparallel to  WF  . Then  BX   =  WF   = 3, XF   =  BW   = 1, and  XS   = 3 follows.Therefore,   BXS   is an isosceles right-angledtriangle, and so  ∠ BSX   = 45 ◦ . 213331 T W  F S  B X  Let  Y    be the point on  SF   such that  TY    isparallel to  WF  . Then  TY    =  WF   = 3, YF   =  TW   = 3, and  YS   = 1 follows.  TYS   is right-angled and therefore, by thePythagorean Theorem, TS  2 =  TY   2 + SY   2 = 3 2 + 1 2 = 10 and so TS   = √  10, since  TS >  0. 33331 Y T W  F S  Draw in  TF  .   TWF   is right angled and so by the Pythagorean Theorem, TF  2 =  TW  2 + WF  2 = 3 2 + 3 2 = 18 and so  TF   = √  18, since  TF >  0.Now we will use the cosine law in   TSF  . TF  2 =  TS  2 + SF  2 − 2( TS  )( SF  )cos( ∠ TSF  )18 = 10 + 4 2 − 2( √  10)(4)cos( ∠ TSF  )18 − 10 − 16 = − 8 √  10cos( ∠ TSF  ) − 8 = − 8 √  10cos( ∠ TSF  )1 √  10= cos( ∠ TSF  ) ∠ TSF   ≈ 71 . 6 ◦ 3341810 T W  F S  Thus,  ∠ TSB  =  ∠ TSF   − ∠ BSX   ≈ 71 . 6 ◦ − 45 ◦  ˙= 26 . 6 ◦ .
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