+9466 | 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 |
+9466 | 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 |
