Question

Из одной точки проведены к плоскости две наклонные проекции которых равны 4,5 и 1,5 дециметра Найдите длины наклонных если 1 из них образует с плоскостью угол 2 раза больше чем другая наклонная

Answer 1

Вот
Расположим обе наклонных в одной вертикальной плоскости, для удобства построения.Точку из которой проведены наклонные обозначим К. Опустим из К перпендикуляр на плоскость до пересечения в точке С. Для удобства примем КС параллельно оси Y.  Из точки С проводим горизонталь АС. Угол АСК прямой. АС=4,5,   ВС=1,5. Обозначим КАС=а, тогда из условия КВС=2а. По известной формуле tg2а=2tgа/(1-tgа квадрат). КС=АСtgа=4,5 tgа. Из второго треугольника КС=ВСtg2а=(1,5 на 2tgа)/(1-tgа квадрат). Отсюда tgа=0,578. Угол а=30. Тогда искомые длины наклонных АК=АС/cosа=5,2   ВК=ВС/cos2а=3.

Answer 2

Problem Analysis

We are given two inclined projections from a point to a plane, with lengths of 4.5 and 1.5 decimeters. We need to find the lengths of the inclined projections if one of them forms an angle with the plane that is twice as large as the other inclined projection.

Solution

Let's assume the lengths of the inclined projections are x and y decimeters, with x being the larger length. We are given that one of the inclined projections forms an angle with the plane that is twice as large as the other inclined projection.

From the given information, we can set up the following equation:

2y = x

Now, we can solve the equation to find the values of x and y.

Calculation

Substituting the value of x from the equation into the equation itself, we get:

2y = 2y

This equation is true for any value of y. Therefore, there are infinitely many possible solutions for the lengths of the inclined projections.

For example, if we choose y = 1.5 decimeters, then x = 3 decimeters. This satisfies the condition that one of the inclined projections forms an angle with the plane that is twice as large as the other inclined projection.

Similarly, if we choose y = 3 decimeters, then x = 6 decimeters. This also satisfies the condition.

In general, for any value of y, we can find the corresponding value of x by multiplying y by 2.

Therefore, the lengths of the inclined projections can vary depending on the chosen value of y.

Answer

The lengths of the inclined projections can vary depending on the chosen value of y. One possible solution is x = 3 decimeters and y = 1.5 decimeters, where x is twice as large as y. Another possible solution is x = 6 decimeters and y = 3 decimeters.

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