# Trig Ratios Right Triangles

• Jul 28th 2007, 05:02 PM
Sarah32
Trig Ratios Right Triangles
I was doing some homework in my math book and I'm having a hard time with these questions. Here is one that i don't have much idea to do with. Grade 11 University math thanks!

Question 1:
The arm of a boom crane is 12m long. Because of the location of the construction site, the angle of inclination of the boom of the crane has a minimum value of 30 degrees and a maximum of 45 degrees. Find the vertical displacement of the end of the boom as
a) an exact value
b) an approximate value, to the nearest tenth of a metre.
• Jul 28th 2007, 06:48 PM
Soroban
Hello, Sarah!

Quote:

1) The arm of a boom crane is 12m long.
Because of the location of the construction site, the angle of inclination
of the boom of the crane has a minimum value of 30° and a maximum of 45°.

Find the vertical displacement of the end of the boom as
a) an exact value
b) an approximate value, to the nearest tenth of a metre.

Code:

                        *                     *  |             12  *    |               *        | y1             *          |         * 30°          |       * - - - - - - - - *
You're expected to the ratio of the sides of a 30-60 right triangle.

The side opposite the 30°-angle is always one-half the hypotenuse.

Hence: . $(b)\;y_1 \,=\,6$ m.

Code:

                  *                 * |           12  *  |             *    | y2           *      |         * 45°    |       * - - - - - *
You should know that sides of a 45-45 right triangle are in the ratio: . $1:1:\sqrt{2}$

Hence: . $y_2\:=\:\frac{12}{\sqrt{2}}\:=\:6\sqrt{2}$

Therefore: . $(a)\;y_2 - y_1 \;=\;6\sqrt{2} - 6 \;=\;6(1-\sqrt{2})\;m$

. . . . . . . . $(b)\;y_2-y_1 \;=\;2.485281376 \;\approx\; 2.5\;m$

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