321 (2) 1.xの値(辺の長さ)を求めよう。 30 30° x 72 = √√√5 x = "( 14 12+√√3 √√3+√√3 = 3 = 4√3 114 28° X X 2 48 √√√√ √√√3 0.8839 (3) 6-88-9 3 3²3²1²6 2 = 123606 088 29 10 (4) 10 45° = 0.7071 x = 7.07/ 123606 V63° X x 20 t ・10 BO 20 20 9= 0.5095 t=10,1900 639-X (5) x 0.5095 20 (6) 160° LANG x = 2√56²₂0 №o = 12 ooooo 0101900 1. √6 130° A 3 7²√3 x №o 3 1 x√√3 * = 3√3+√√√3 E g λ = x M 3 (7) 8 14 (8) x=0.2419 八=1.9352 0.2℃ 12352 352 x O 3√2/30° x X 8 x 3√> √√3x √√2 d = futzatz NG 12 3√2