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 Oct 14, 2018

Best Answer 

 #1
avatar+7352 
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m∠ABE  =  m∠CBE__because parallelogram ABEF  ≅  parallelogram CBED .
m∠ABC  =  m∠ABE + m∠CBE by the angle addition postulate.
30°  =  m∠ABE + m∠ABE by substitution.
30°  =  2m∠ABE  
15°  =  m∠ABE  

 

Let's draw a height of parallelogram ABEF so that BE and AF are the bases, and call it  h .

 

 

sin( m∠ABE )  =  h / AB
sin( 15° )  =  h / x

x sin 15°  =  h

 

 

sin( m∠PBE )  =  h / BP

sin( m∠PBE )  =  x sin 15° / 10

m∠PBE  =  arcsin( x sin 15° / 10 )

 

And  m∠PBE  =  (m∠PBQ)/2

 

(m∠PBQ)/2  =  arcsin( x sin 15° / 10 )__ 

m∠PBQ  =  2 arcsin( x sin 15° / 10 )

 

 
cos( m∠PBQ )  =  cos( 2 arcsin( x sin 15° / 10 ) )

 

 

cos( m∠PBQ )  =  1 - 2[ x sin 15° / 10 ]2 

 

By the double angle formula for cosine:

cos(2u)  =  1 - 2 sin2u

cos( m∠PBQ )  =  1 - 2 x2 sin215° / 100

 

 
cos( m∠PBQ )  =  1 - 2 x2 sin2(30/2°) / 100

 

 
cos( m∠PBQ )  =  1 - 2 x2 ( (1 - cos 30°) / 2 ) / 100 

By the half-angle formula for sine:

sin2( u/2 )  =  (1 - cos u) / 2

cos( m∠PBQ )  =  1 - x2 (1 - cos 30°) / 100

 

 
cos( m∠PBQ )  =  1 - x2 (1 - √3 / 2) / 100

 

 
cos( m∠PBQ )  =  1 - x2 (2 - √3) / 200  
 Oct 15, 2018
 #1
avatar+7352 
+1
Best Answer
m∠ABE  =  m∠CBE__because parallelogram ABEF  ≅  parallelogram CBED .
m∠ABC  =  m∠ABE + m∠CBE by the angle addition postulate.
30°  =  m∠ABE + m∠ABE by substitution.
30°  =  2m∠ABE  
15°  =  m∠ABE  

 

Let's draw a height of parallelogram ABEF so that BE and AF are the bases, and call it  h .

 

 

sin( m∠ABE )  =  h / AB
sin( 15° )  =  h / x

x sin 15°  =  h

 

 

sin( m∠PBE )  =  h / BP

sin( m∠PBE )  =  x sin 15° / 10

m∠PBE  =  arcsin( x sin 15° / 10 )

 

And  m∠PBE  =  (m∠PBQ)/2

 

(m∠PBQ)/2  =  arcsin( x sin 15° / 10 )__ 

m∠PBQ  =  2 arcsin( x sin 15° / 10 )

 

 
cos( m∠PBQ )  =  cos( 2 arcsin( x sin 15° / 10 ) )

 

 

cos( m∠PBQ )  =  1 - 2[ x sin 15° / 10 ]2 

 

By the double angle formula for cosine:

cos(2u)  =  1 - 2 sin2u

cos( m∠PBQ )  =  1 - 2 x2 sin215° / 100

 

 
cos( m∠PBQ )  =  1 - 2 x2 sin2(30/2°) / 100

 

 
cos( m∠PBQ )  =  1 - 2 x2 ( (1 - cos 30°) / 2 ) / 100 

By the half-angle formula for sine:

sin2( u/2 )  =  (1 - cos u) / 2

cos( m∠PBQ )  =  1 - x2 (1 - cos 30°) / 100

 

 
cos( m∠PBQ )  =  1 - x2 (1 - √3 / 2) / 100

 

 
cos( m∠PBQ )  =  1 - x2 (2 - √3) / 200  
hectictar Oct 15, 2018
 #2
avatar+98153 
+1

Excellent, hectictar  !!!

 

cool cool cool

CPhill  Oct 15, 2018

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