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Two support cables are attached on opposite sides of a tower Phone portable. One of...

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misztibabe | eNoter

Posted July 8, 2013 at 2:17 AM via web

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Two support cables are attached on opposite sides of a tower Phone portable. One of the cables is attached to the floor at a point A 9 m from the base of the tower. The second cable is attached to B at an angle of elevation of 30° and meeting the tower same location as the first cable. The attachment point at ground level are located at a distance of 17 m. What is the minimum amount of wiring required to support  them both?

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Mary Joy Ripalda | High School Teacher | (Level 3) Educator

Posted July 8, 2013 at 3:25 AM (Answer #1)

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Refer to the figure below which illustrates the given in the problem.

In the figure below, notice that that the length of the second cable can be readily solved using the triangle CDB.

In the this triangle, angle B and its adjacent side are known. So the hypotenuse CB can be solved using the formula of cosine.

`cos B=(adjacent)/(hypoten u se)`

`cos B=(DB)/(CB)`

`cos 30^o=17/(CB)`

`CB=17/(cos30^o)`

CB=19.63

Hence, the length of the second cable is 19.63 m.

To solve for the length of the first cable CA, consider the triangle ACB.

Since the length of its two sides (AB and CB) and its included angle (B) are known, apply Cosine Law.

`(AC)^2=(AB)^2+(CB)^2-2(AB)(CB)cosB`

Then, plug-in AB=9+17=26, CB=19.63 and B=30.

`(AC)^2=26^2+19.63^2-2*26*19.63cos30^o`

`(AC)^2=177.33`

`AC=sqrt177.33`

`AC=13.32`

Thus, the length of the first cable is 13.32 m.

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