1. Т.к. BCA - тупой => ∠BAC и ∠ABC- острые
2. Т.к. ∠KBC меньше ∠ABC ,а ∠BCA = ∠BCK, то ∠BAC меньше ∠BKC => sin(BKC) > sin(BAC)
3. По теореме синусов: BC/(sin(BKC)) = BK/sin(150) и BC/sin(BAC) = AB/sin(150), =>
=> = sin(BKC)/sin(BAC) => AB > BK
4. Т.к. 6 > AB > BK = 4,5 и AB - целое, то AB = 5.
Ответ: AB = 5.
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Answers & Comments
1. Т.к. BCA - тупой => ∠BAC и ∠ABC- острые
2. Т.к. ∠KBC меньше ∠ABC ,а ∠BCA = ∠BCK, то ∠BAC меньше ∠BKC => sin(BKC) > sin(BAC)
3. По теореме синусов: BC/(sin(BKC)) = BK/sin(150) и BC/sin(BAC) = AB/sin(150), =>
=> = sin(BKC)/sin(BAC) => AB > BK
4. Т.к. 6 > AB > BK = 4,5 и AB - целое, то AB = 5.
Ответ: AB = 5.