trigo Q(2)

Required ∠s = ∠BPKand ∠BQK = ∠BPK= ∠BPD - Φ= tan⁻¹(DB/10)- tan⁻¹(DK/10)= tan⁻¹[√(9² +12²)/10] - tan⁻¹{[√(9² +12²) – 5]/10}= tan⁻¹(3/2)- tan⁻¹1= tan⁻¹(3/2)- 45°≈ 11.3° = ∠BQK= sin⁻¹(5/BQ)= sin⁻¹[5/√(9² +10²)]= sin⁻¹(5/√181)≈ 21.8° I HOPETHIS CAN HELP YOU! ~ ^_^

2012-07-28 12:22:13 補充：
Wait... There's something wrong with ∠BQK.
Let me correct it...

2012-07-28 13:10:06 補充：
Here's the correct ans of ∠BQK
http://i1099.photobucket.com/albums/g395/jasoncube/16c0003a.png

Note that sin⁻¹(5/√181) ≈ 21.817° and cos⁻¹(77/√6878) ≈ 21.805°.
I was wrong before because I thought ∆BKQ is a right - angled ∆,
but in fact it's not.

Sorry for my mistake ~

angle rotated by P=∠BPK=α

tanφ=DK/PD=10/10=1⇒φ=45°
tan(φ+α)=BD/PD=15/10⇒φ+α=56.3°

∴α=11.3°

angle rotated by Q=∠BQK=β

In ∆BDQ, cos∠BDQ=(DB²+DQ²-BQ²)/(2×DB×DQ)
=(15²+(10²+12²)-(10²+9²))/(2×15×√(10²+12²))=288/(60√61)=0.6146

In ∆KDQ, KQ²=DQ²+DK²-2×DQ×DK×cos∠BDQ
=(10²+12²)+10²-2×√(10²+12²))×10×0.6146=151.99

In ∆KBQ, cosβ=(KQ²+BQ²-BK²)/(2×KQ×BQ)=(151.99+181-25)/(2×√151.99×√181)=0.9285

∴β=21.8°