長方形OAB$_1$C$_1$において$\text{OA}=1,\ \angle \text{AOB}_1=\theta \ (0^\circ<\theta<90^\circ)$とする．図のように，この長方形の対角線OB$_1$を一辺とし，$\angle \text{B}_1 \text{OB}_2=\theta$となる長方形OB$_1$B$_2$C$_2$を反時計回りに作る．同様にして$\angle \text{B}_n \text{OB}_{n+1}=\theta$となる長方形OB$_n$B$_{n+1}$C$_{n+1} \ (n=1,\ 2,\ \cdots)$を作る．次の問いに答えよ．
(1) 線分OB$_1$およびB$_1$B$_2$の長さを$\theta$で表せ．
(2) 長方形OB$_n$B$_{n+1}$C$_{n+1}$の面積を$n$と$\theta$で表せ．ただしB$_0=\text{A}$とする．
(3) $\theta=30^\circ$のとき，図形OAB$_1$B$_2$B$_3$B$_4$C$_4$の面積$S$を求めよ．
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