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zhbr991
@zhbr991
July 2022
1
12
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В треугольнике авс стороны ас и бс равны, аб равно 12, cos а равен 2√5/5. Найти высоту сн
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mashuliako1
AB=AC/sinA
sinA=sqrt(1-cos^2A)=sqrt(1-16/25)=3/5
AB=3/(3/5)=5
AC=sqrt(25-9)=4
AH=AC*cosA=4*cosA=4*4/5=16/5=3.2
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zhbr991
а где взял 25?
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Answers & Comments
sinA=sqrt(1-cos^2A)=sqrt(1-16/25)=3/5
AB=3/(3/5)=5
AC=sqrt(25-9)=4
AH=AC*cosA=4*cosA=4*4/5=16/5=3.2