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TTPOFECCOP1
@TTPOFECCOP1
July 2022
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Найдите длины сторон и градусные меры углов ABD и BDC.
СРОЧНООО
ДАЮ 40 БАЛЛОВ!!!!
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alex57school
Угол ABD = 180 - 90 - 41 = 49
BD = AD * tg(41) = 0.5 * tg(41)
AB = AD / cos(41) = 0.5 / cos(41)
В треугольнике DCB:
1.8^2 = BD^2 + CD^2
1.8^2 = 0.25 * tg^(41) + CD^2
CD^2 = 1.8^2 - 0.25 * tg^2(41)
CD = sqrt(1.8^2 - 0.25 * tg^2(41))
sin(уголBCD) = BD / BC = 0.5 * tg(41) / 1.8
уголBCD = arcsin(0.5 * tg(41) / 1.8)
sin(уголDBC) = DC / BC = sqrt(1.8^2 - 0.25 * tg^2(41)) / 1.8
уголDBC = arcsin(sqrt(1.8^2 - 0.25 * tg^2(41)) / 1.8)
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Answers & Comments
BD = AD * tg(41) = 0.5 * tg(41)
AB = AD / cos(41) = 0.5 / cos(41)
В треугольнике DCB:
1.8^2 = BD^2 + CD^2
1.8^2 = 0.25 * tg^(41) + CD^2
CD^2 = 1.8^2 - 0.25 * tg^2(41)
CD = sqrt(1.8^2 - 0.25 * tg^2(41))
sin(уголBCD) = BD / BC = 0.5 * tg(41) / 1.8
уголBCD = arcsin(0.5 * tg(41) / 1.8)
sin(уголDBC) = DC / BC = sqrt(1.8^2 - 0.25 * tg^2(41)) / 1.8
уголDBC = arcsin(sqrt(1.8^2 - 0.25 * tg^2(41)) / 1.8)