BH⊥AD; BH = 8·sin(30) = 4; KD = AD = 8; KH = KD·sin(60) = 4√3
d(k, BC) = BK = 4√2
BK² = BH² + KH² - 2·BH·KH·cos(∠KHB)
cos(∠KHB) = (√3)/3
Объяснение:
BH⊥AD; BH = 8·sin(30) = 4; KD = AD = 8; KH = KD·sin(60) = 4√3
d(k, BC) = BK = 4√2
BK² = BH² + KH² - 2·BH·KH·cos(∠KHB)
cos(∠KHB) = (√3)/3
Объяснение: