Question

A tower stands vertically on the ground. From a point which is 15 meter away from the
foot of the tower, the angle of elevation of the top of the tower is 45°. What is the height
of the tower?
​

Answer 1

Answer:

The answer is 15m.

Step-by-step explanation:

The method is given in the attachment.

Answer 2

ANSWER:

Given:

• A point is taken 15m away from ground.
• Angle of elevation = 45°

To Find:

• Height of the tower

Diagram:

$\setlength{\unitlength}{1cm}\begin{picture}(6,5)\linethickness{.4mm}\put(1,1){\line(1,0){4.5}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(.3,2.5){\large\bf h}\put(2.8,.3){\large\bf 15m}\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf A}\put(.8,.3){\large\bf B}\put(5.8,.3){\large\bf C}\qbezier(4.5,1)(4.3,1.25)(4.6,1.7)\put(3.8,1.3){\large\bf 45^{\circ} }\end{picture}$

Solution:

In the given diagram, AB is the tower.

So,

$\sf{:\longrightarrow \tan\theta = \dfrac{perpendicular}{base}} \\\\\sf{:\implies \tan{45^{\circ}} = \dfrac{AB}{BC}} \\\\\sf{:\implies 1 = \dfrac{h}{15m}} \\\\\bf{:\implies h = 15m}$

Therefore, the height of the tower is 15m.

Formulae Used:

• tanΘ = perpendicular/base
• tan 45° = 1

Learn More:

TRIGONOMETRIC VALUES:

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

TRIGONOMETRIC RATIOS:

• sinΘ = perpendicular/hypotenuse
• cosΘ = base/hypotenuse
• tanΘ = perpendicular/base
• cotΘ = base/perpendicular
• secΘ = hypotenuse/base
• cosecΘ = hypotenuse/perpendicular