6 loads of stone weigh 2/3 ton. Find the weight of 1 load of stone.

1 answer:

4 0

$\bf \begin{array}{ccll} loads&tons\\ \cline{1-2} 6&\frac{2}{3}\\\\ 1&x \end{array}\implies \cfrac{6}{1}=\cfrac{~~\frac{2}{3}~~}{x}\implies 6=\cfrac{~~\frac{2}{3}~~}{\frac{x}{1}}\implies 6=\cfrac{2}{3}\cdot \cfrac{1}{x} \\\\\\ 6=\cfrac{2}{3x}\implies 18x=2\implies x=\cfrac{2}{18}\implies x=\cfrac{1}{9}$

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Step-by-step explanation:

The triangle in the given problem is a right triangle, as the tower forms a right angle with the ground. This means that one can use the right angle trigonometric ratios to solve this problem. The right angle trigonometric ratios are as follows;

$sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}$

Please note that the names ( $opposite$ ) and ( $adjacent$ ) are subjective and change depending on the angle one uses in the ratio. However the name ( $hypotenuse$ ) refers to the side opposite the right angle, and thus it doesn't change depending on the reference angle.

In this problem, one is given an angle with the measure of (35) degrees, and the length of the side adjacent to this angle. One is asked to find the length of the side opposite the (35) degree angle. To achieve this, one can use the tangent ( $tan$ ) ratio.