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A flagstaff, $7.32$ m long is fitted at the top of $10$ m tall building. The flagstaff is supported by the ropes which are tied to the point P on the ground which is $x$ m away from the base of the building. It is given that $l_1$ is the length of rope from point P to the base of the flagstaff and $l_2$ is the length of rope from point P to the top of flagstaff. Rope $l_1$ makes an angle of $30^\circ$ with the horizontal and $\theta$ be the angle which rope $l_2$ makes with the horizontal as shown in the figure.
Based on the above information, answer the following questions :
(Use $\sqrt{2} = 1.4$ and $\sqrt{3}= 1.732$)
(i) Find the value of $x$.
(ii) Find the measure of angle $\theta$.
(iii) (a) Find the total length of ropes needed to support the flagstaff.
OR
(iii) (b) Which rope is longer $l_1$ or $l_2$ and by how much ?
Based on the above information, answer the following questions :
(Use $\sqrt{2} = 1.4$ and $\sqrt{3}= 1.732$)
(i) Find the value of $x$.
(ii) Find the measure of angle $\theta$.
(iii) (a) Find the total length of ropes needed to support the flagstaff.
OR
(iii) (b) Which rope is longer $l_1$ or $l_2$ and by how much ?
Show SolutionHide Solution↓
Ans. (i) $\frac{10}{x} = \tan 30^\circ$ (1/2 Mark)
$x = 17.32$ m (1/2 Mark)
(ii) $\frac{17.32}{17.32} = \tan \theta$ (1/2 Mark)
$\theta = 45^\circ$ (1/2 Mark)
(iii) (a) $\frac{10}{l_1} = \sin 30^\circ$ (1/2 Mark)
gives $l_1 = 20$ m (1/2 Mark)
Similarly, $l_2 = 17.32 \sqrt{2} = 24.248$ m (1/2 Mark)
Total length of rope needed $= 44.248$ m (1/2 Mark)
OR
(iii) (b) $\frac{10}{l_1} = \sin 30^\circ$ (1/2 Mark)
Gives $l_1 = 20$ m (1/2 Mark)
Similarly, $l_2 = 17.32 \sqrt{2} = 24.248$ m (1/2 Mark)
$l_2$ is longer than $l_1$ by $4.248$ m (1/2 Mark)
$x = 17.32$ m (1/2 Mark)
(ii) $\frac{17.32}{17.32} = \tan \theta$ (1/2 Mark)
$\theta = 45^\circ$ (1/2 Mark)
(iii) (a) $\frac{10}{l_1} = \sin 30^\circ$ (1/2 Mark)
gives $l_1 = 20$ m (1/2 Mark)
Similarly, $l_2 = 17.32 \sqrt{2} = 24.248$ m (1/2 Mark)
Total length of rope needed $= 44.248$ m (1/2 Mark)
OR
(iii) (b) $\frac{10}{l_1} = \sin 30^\circ$ (1/2 Mark)
Gives $l_1 = 20$ m (1/2 Mark)
Similarly, $l_2 = 17.32 \sqrt{2} = 24.248$ m (1/2 Mark)
$l_2$ is longer than $l_1$ by $4.248$ m (1/2 Mark)