KnowledgeBoat Logo
|

Mathematics

From the top of a pillar of height 20 m, the angles of elevation and depression of the top and bottom of another pillar are 30° and 45° respectively. The height of the second pillar (in metres) is :

  1. (20(31)3)\Big(\dfrac{20(\sqrt{3}-1)}{\sqrt{3}}\Big)

  2. 10

  3. 10310\sqrt{3}

  4. (203(3+1))\Big(\dfrac{20}{\sqrt{3}}(\sqrt{3}+1)\Big)

Heights & Distances

1 Like

Answer

Let AB and CD be two pillars.

Draw a line from A to meet CD at point E, AE = x

From the top of a pillar of height 20 m, the angles of elevation and depression of the top and bottom of another pillar are 30° and 45° respectively. The height of the second pillar (in metres) is : Volume And Surface Area of solid RSA Mathematics Solutions ICSE Class 10.

In triangle ADE,

tan45=EDAE1=20xx=20 m.\Rightarrow \tan 45^{\circ} = \dfrac{ED}{AE} \\[1em] \Rightarrow 1 = \dfrac{20}{x} \\[1em] \Rightarrow x = 20 \text{ m.}

In triangle ACE,

tan30=CEAE13=CE20CE=203 m.\Rightarrow \tan 30^{\circ} = \dfrac{CE}{AE} \\[1em] \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{CE}{20} \\[1em] \Rightarrow CE = \dfrac{20}{\sqrt3} \text{ m.}

The total height of the second pillar is,

CD = ED + CE

CD=20+203CD=20(1+13)CD=20(3+13)CD=203(3+1) m.\Rightarrow CD = 20 + \dfrac{20}{\sqrt{3}} \\[1em] \Rightarrow CD = 20 \Big(1 + \dfrac{1}{\sqrt{3}} \Big) \\[1em] \Rightarrow CD = 20 \Big( \dfrac{\sqrt{3} + 1}{\sqrt{3}} \Big) \\[1em] \Rightarrow CD = \dfrac{20}{\sqrt{3}}(\sqrt{3} + 1) \text{ m}.

Hence, option 4 is the correct option.

Answered By

2 Likes


Related Questions