PROBLEM-SOLVING $\newline$ $6$ A $100 \mathrm{~m}$ radio mast is supported by six cables in two sets of three cables. All six cables are anchored to the ground at an equal distance from the mast. The top set of three cables is attached at a point $20 \mathrm{~m}$ below the top of the mast. Each of the three lower cables is $60 \mathrm{~m}$ long and attached at a height of $30 \mathrm{~m}$ above the ground. If all the cables have to be replaced, find the total length of cable required. Give your answer correct to two decimal places.

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Q. PROBLEM-SOLVING

\newline

6

100 \mathrm{~m}

radio mast is supported by six cables in two sets of three cables. All six cables are anchored to the ground at an equal distance from the mast. The top set of three cables is attached at a point

20 \mathrm{~m}

below the top of the mast. Each of the three lower cables is

60 \mathrm{~m}

long and attached at a height of

30 \mathrm{~m}

above the ground. If all the cables have to be replaced, find the total length of cable required. Give your answer correct to two decimal places.

Calculate cable length: Calculate the length of one of the top set of cables using the Pythagorean theorem. The mast is $100\,\text{m}$ tall, and the top cables are attached $20\,\text{m}$ below the top, so they are attached at a height of $80\,\text{m}$ . The bottom of the mast is $0\,\text{m}$ , so the vertical distance from the ground to the attachment point is $80\,\text{m}$ . Since the cables are anchored at the same distance from the mast and the lower cables are $60\,\text{m}$ long, we can assume the horizontal distance is the same for the top cables. We can use the Pythagorean theorem to find the length of one top cable: $a^2 + b^2 = c^2$ , where $a$ is the vertical distance ( $80\,\text{m}$ ), $b$ is the horizontal distance, and $20\,\text{m}$ $0$ is the length of the cable. We already know that $20\,\text{m}$ $0$ (the length of the lower cable) is $60\,\text{m}$ .
Use Pythagorean theorem: Now, we need to solve for $b$ , the horizontal distance. We have $c = 60\,\text{m}$ and $a = 80\,\text{m}$ . Plugging these into the Pythagorean theorem: $80^2 + b^2 = 60^2$ .