Question

A pole is 12 m high, A steel wire tied to the top of the pole is affixed at a point on the ground to keep the pole upright. The steel wire makes an angle of 60° with the horizontal through the foot of the pole. Find the length of the steel wire to the nearest meter.( √3=1.73)​

Answer 1

Given :

• Height of the Pole = 12 m.

• Angle of Elevation = 60°

To find :

The length of the wire

Solution :

According to the given information, we are provided with the height of the figure (here triangle) and the angle of Elevation.

And according to the figure (Right-angled triangle) , AB is the height of the triangle and AC is the Hypotenuse of the triangle , and height with hypotenuse is taken as sin θ , since we have to find the Hypotenuse.i.e,

The length of the wire is the Hypotenuse of the triangle.

We that sin θ is :-

$\boxed{\bf{\sin\:\theta = \dfrac{P}{H}}}$

Where :

• P = Height of the triangle
• H = Hypotenuse of the triangle

So, using sin θ and substituting the given values in it, we get :

$:\implies \bf{\sin\:60^{\circ} = \dfrac{12}{H}}$

$:\implies \bf{\dfrac{\sqrt{3}}{2} = \dfrac{12}{H}}\quad [\because \bf{sin\:60^{\circ} = \dfrac{\sqrt{3}}{2}}]$

$:\implies \bf{\dfrac{1.73}{2} = \dfrac{12}{H}} \quad [\because \bf{\sqrt{3} = 1.73}]$

$:\implies \bf{1.73 = \dfrac{12}{H} \times 2}$

$:\implies \bf{1.73 = \dfrac{24}{H}}$

$:\implies \bf{H = \dfrac{24}{1.73}}$

$:\implies \bf{H = 13.9(approx.)}$

$\boxed{\therefore \bf{Hypotenuse\:(H) = 13.9\:m}}$

Hence, the Hypotenuse of the triangle is 13.9 m.

Since, we have taken the length of the wire as the Hypotenuse , the length of the wire is 13.9 m.

Alternative method :

Solution :

To find the hypotenuse of the triangle , first we have to find the base of the triangle and then by using the Pythagoras theorem , we can find the required value.

To find the base of the triangle :

Using tan θ and substituting the values in it, we get :

$\boxed{\bf{\sin\:\theta = \dfrac{P}{B}}}$

Where :

• P = Height of the triangle
• B = Base of the triangle

$:\implies \bf{\tan\:60^{\circ} = \dfrac{12}{B}}$

$:\implies \bf{\sqrt{3} = \dfrac{12}{B}}\quad [\because \bf{tan\:60^{\circ} = \sqrt{3}}]$

$:\implies \bf{1.73 = \dfrac{12}{B}} \quad [\because \bf{\sqrt{3} = 1.73}]$

$:\implies \bf{B = \dfrac{12}{1.73}} \quad [\because \bf{\sqrt{3} = 1.73}]$

$:\implies \bf{B = 6.9}$

$\boxed{\therefore \bf{Base\:(B) = 6.9\:m}}$

Hence, the base of the triangle is 6.9 m.

To find the Hypotenuse of the triangle :

Here,

• B = 6.9 m
• P = 12 m.

By using the Pythagoras theorem and substituting the values in it, we get :

$\boxed{\bf{H^{2} = P^{2} + B^{2}}}$

Where :

• B = Base of the triangle
• H = Hypotenuse of the triangle
• P = Height of the triangle

$:\implies \bf{H^{2} = 12^{2} + 6.9^{2}}$

$:\implies \bf{H = \sqrt{12^{2} + 6.9^{2}}}$

$:\implies \bf{H = \sqrt{144 + 47.6}}$

$:\implies \bf{H = \sqrt{191.6}}$

$:\implies \bf{H = 13.9}$

$\boxed{\therefore \bf{Hypotenuse\:(H) = 13.9\:m}}$

Hence, the Hypotenuse of the triangle is 13.9 m.

Since, we have taken the length of the wire as the Hypotenuse , the length of the wire is 13.9 m.

Answer 2

$\setlength{\unitlength}{1cm}\begin{picture}(6,6)\put(2, 2){\line(2,1){2}}\put(2,2){\line(1,0){2}}\put(4,2){\line(0,2){1}}\end{picture}$

$\large\sf \pink{Given} \purple{ \begin{cases} \red{ \sf\mapsto Height \: of \: pole = 12m } \\ \\ \orange{ \sf \mapsto Angle \: of \: elevation = {60}^{ \circ} } \end{cases}}$

$\large\sf \purple{Find} \orange{ \begin{cases} \\ \blue{ \sf\mapsto Length \: of \: steel \: wire } \\ \\ \end{cases}}$

$\large\sf \purple{Solution}$

Here,

$\sf \hookrightarrow\dfrac{AC}{AB} = \dfrac{H}{P}$

$\sf \hookrightarrow \dfrac{H}{P} = \cosec {60}^{ \circ}$

$\sf \hookrightarrow \dfrac{AC}{AB} = \cosec {60}^{ \circ}$

$\sf \hookrightarrow \dfrac{AC}{AB} = \dfrac{2}{ \sqrt{3} } \qquad \bigg\{ \because \cosec {30}^{ \circ} = \dfrac{2}{ \sqrt{3} } \bigg\}$

$\sf \hookrightarrow \dfrac{AC}{12} = \dfrac{2}{ \sqrt{3} } \qquad \bigg\{ \because AB = 12m\bigg\}$

Now, Do Cross-multiplication

$\sf \hookrightarrow AC \sqrt{3} = 12 \times 2$

$\sf \hookrightarrow AC \sqrt{3} = 24$

$\sf \hookrightarrow AC = \dfrac{24}{ \sqrt{3}}$

$\sf \hookrightarrow AC = \dfrac{24}{1.73} \qquad \{ given \: \sqrt{3} = 1.73\}$

$\sf \hookrightarrow AC = \dfrac{24}{1.73} = 13.872(approx.)$

$\sf \hookrightarrow AC =13.9m$

Hence, Height of the steel wire is 13.9m