第11章 11.16

(1)

$\tan\theta = \dfrac{94}{188 \div 2} = \dfrac{94}{94} = 1 \qquad \theta = 45°$

(2)

底辺中心から底辺の角までの距離$94\sqrt{2}$はなので

$\tan\theta = \dfrac{94}{94\sqrt{2}} = \dfrac{1}{\sqrt{2}} \qquad \theta \cong 35.26°$

(3)

側面の二等辺三角形の高さは$94\sqrt{2}$なので

$\tan\dfrac{\theta}{2} = \dfrac{94}{94\sqrt{2}} = \dfrac{1}{\sqrt{2}} \qquad \dfrac{\theta}{2} \cong 35.26 \qquad \theta = 70.5°$