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1) В правильной треугольной пирамиде проекции боковых рёбер на основание имеют угол между собой в 360°/3 = 120°.
Тангенс угла β наклона бокового ребра к основанию равен:
tg β = tg α*cos φ = tg 47*cos(120/2) = 1,07236871 * 0,5 = 0,536184355.
β = arc tg(0,536184355) = 0,492174352 радиан = 28,19951313°.
2) Проведём осевое сечение пирамиды через апофему.
В сечении - равносторонний треугольник.
Высота из середины основания этого треугольника на боковую сторону равна 2 см ( по заданию - это расстояние от центра основания до боковой грани).
Высота Н пирамиды как гипотенуза в 2 раза больше катета, лежащего против угла в 30°: Н = 2*2 = 4 см.
Апофема А равна стороне а основания: A = а.
По Пифагору А² = (а/2)² + Н²,
а² = (а/2)² + 4².
4а² = а² + 16*4,
3а² = 64,
а = √(64/3) = 8/√3 = 8√3/3 см.
Периметр основания Р = 4а = 4*(8√3/3) = 32√3/3 см.
Искомая площадь боковой поверхности пирамиды равна:
Sбок = (1/2)РА = (1/2)*(32√3/3)*(8√3/3) =128/3 = 42(2/3) см².
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