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# Please explain this question, i dont know where to start!

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Oct 14, 2018

#1
+7347
+1
 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
+7347
+1
 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
+94321
+1

Excellent, hectictar  !!!

CPhill  Oct 15, 2018