The supporting wire to the top of a vertical pole is 17 m long is fastened to the ground 8 m from the foot of the pole, then how high is the pole? ( in m )

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Solution

Consider the structure of pole and supporting wire as a right-angle triangle, with the base as 8 m and the length of supporting wire as the hypotenuse and length of pole itself being the perpendicular.
Thus, $H^{2} = P^{2} + B^{2}$
$17^{2} = 8^{2} + P^{2}$
$P^{2} = 225$
$P = 15$ m
Thus, length of the pole $= 15 m$

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The supporting wire to the top of a vertical pole is 17 m long is fastened to the ground 8 m from the foot of the pole, then how high is the pole? ( in m )

Consider the structure of pole and supporting wire as a right-angle triangle, with the base as 8 m and the length of supporting wire as the hypotenuse and length of pole itself being the perpendicular.Thus, H2=P2+B2172=82+P2P2=225P=15 mThus, length of the pole =15m