Question

Two vertical buildings of heights 30 m and 37 m are 16 m apart . A cable is pulled taut and attached from the top of the building to the top of the other building. Find the length of the cable

Answer 1

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Answer 2

The length of the cable is  approximately 17.5 m.

Explanation:

By considering the given description , we made a diagram (given in attachment).

Building with height  37 m is represented by AC .

Building with height  30 m is represented by  ED .

Distance between both building is represented by CD.

Length of the cable  is represented by AE.

Joining E to a point B on AC, we get a rectangle BCDE , such that BC=ED=  30 m and BE=CD= 16 m.

Now , AB= AC-BC= 37-30=7m

We use Pythagoras theorem in right triangle ABE , we get

$AE^2=AB^2+BE^2\\\\ AE^2=7^2+16^2=49+256=305\\\\\Rightarrow\ AE=\sqrt{305}\approx17.5$

Hence, the length of the cable is  approximately 17.5 m.

# Learn more :

A cable wire of length 61m is attached to the top of a building from the bottom of a pole fixed at a distance of 11 m from the bottom of the building. find the height of the building

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