Δ ABC, ∠ C=90, CH- высота , AB=26; tgB=5. AH = ?
Решение
1) 1 + tg²B = 1/Cos²B
1 + 25 = 1/Cos²B
26 = 1/Cos²B
Cos²B = 1/26
CosB = √26/26
2) Sin²B = 1 - Cos²B = 1 - 1/26 = 25/26
SinB = 5√26/26 = CosA
3) CA/AB = CosA
CA/26 = 5√26/26
CA = 5√26
4) AH/CA = CosA
AH = CA* CosA = 5√26* 5√26/26 = 25
5) Ответ: AH = 25
<span>56/36 =14/9 сократили на 4
64/72 =8/9 сократили на 8
14/9+8/9=22/9=2 4/9 сумма
14/9*8/9=112/81=1 31/81 произведение
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