Question

The angle of elevation of the top of a tower, from a point on the ground and at a distance of 150m from its foot, is 30° . Find the height of the tower correct to one place of decimal.​

Answer 1

Answer

The height of the tower is 86.5m

$\bf\large\underline{Given}$

• The angle of elevation of a tower , from a point on the ground and at a distance of 150m from its foot, is 30°

$\bf\large\underline{To \: Find}$

• The height of the tower

$\bf\large\underline{Solution}$

Let us consider the height of the tower be x m

From ‘tan’ function we have :

$\sf\bullet \: \: \tan\theta =\dfrac{height}{base}$

Applying trigonometric ‘tan’ on the given data :

$\sf\implies \tan30\degree = \dfrac{x}{150m} \\\\ \sf\implies \dfrac{1}{\sqrt{3}}=\dfrac{x}{150m}\\\\ \sf\implies x = \dfrac{150m}{\sqrt{3}} \\\\ \sf\implies x = 50\sqrt{3} m \\\\ \sf\implies x = 86.5m$

Thus , height of the tower correct to one place of decimal is 86.5m

Answer 2

$\sf\huge\blue{\underline{\underline{ Question : }}}$

The angle of elevation of the top of a tower, from a point on the ground and at a distance of 150m from its foot, is 30° . Find the height of the tower correct to one place of decimal.

$\sf\huge\blue{\underline{\underline{ Solution : }}}$

★ Given that,

• Angle of elevation of the tower from a point on the ground at a distance of 150 m, with an angle of 30°.

★ To find,

• Height of the tower.

★ According to the Question,

• It forms a right - angled triangle.

◼ ∆ABC is a right - angled triangle.

★ Now,

From the figure, (in the attachment)

• BC = 150 m.
• ∠ ACB = 30°
• Height of the tower = AB = h m.

◼ We know that, the adjacent side and we

need to find the opposite side of ∠ ACB in thr triangle ∆ABC. Hence, we need to consider the trigonometric ratio ' tan ' to solve this problem.

$\sf\:\implies \tan\theta = \frac{AB}{AC}$

• tan 30° = 1/√3

$\sf\:\implies \tan 30 = \frac{h}{150}$

$\sf\:\implies \frac{1}{\sqrt{3}} = \frac{h}{150}$

$\sf\:\implies h\sqrt {3} = 150$

$\sf\:\implies h= \frac{150}{\sqrt{3}}$

• Rationalize the numerator & denominator with√3.

$\sf\:\implies h = \frac{150}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$\sf\:\implies h = \frac{150\sqrt{3}}{3}$

$\sf\:\implies h = 50\sqrt{3}$

• √3 = 1.73

$\sf\:\implies h = 50(1.73)$

$\sf\:\implies h = 86.5$

$\underline{\boxed{\bf{\purple{\therefore The\:height\:of\:the\:tower\:is\:86.5 meters.}}}}\:\orange{\bigstar}$