(1)在△ABC中,AB=2a,∠CAB=θ
所以AC=2acosθ,BC=2asinθ
因为正方形PQRS的边长为x
所以AC=
x |
sinθ |
x |
sinθ |
∴x=
2asin2θ |
2+sin2θ |
(2)因为△ABC中,AC=2acosθ,BC=2asinθ
所以s1=4a2sinθcosθ=2a2sin2θ
因x=
2asin2θ |
2+sin2θ |
所以s2=
4a2(sin2θ)2 |
(2+sin2θ)2 |
因此“规划合理度”
s1 |
s2 |
(2+sin2θ)2 |
2sin2θ |
π |
2 |
s1 |
s2 |
(2+sin2θ)2 |
2sin2θ |
1 |
2 |
4 |
sin2θ |
9 |
2 |
当且仅当sin2θ=1即θ=
π |
4 |
9 |
2 |
(1)如图,在Rt△ABC中,AC=20sinθ,AB=20cosθ,
S1=
1 |
2 |
设正方形的边长为x则BQ=
x |
tanθ |
∴
x |
tanθ |
∴x=
20 | ||
|
20sin2θ |
2+sin2θ |
20sin2θ |
2+sin2θ |
(2)t=sin2θ而S2=
202sin2θ |
4+4sin2θ+sin22θ |
S1 |
S2 |
1 |
4 |
1 |
t |
∵0<θ<
π |
2 |
1 |
4 |
1 |
t |
当t=1时
S1 |
S2 |
3 |
2 |
π |
4 |