If a pole 6 m high casts a shadow 2√3m long on the ground, then the Sun’s elevation is

(a) 60°                      (b) 45°                     (c) 30°                  (d) 90°

-lock-

Slide30.JPG Slide31.JPG

 

-endlock-

You saved atleast 2 minutes of distracting ads by going ad-free. Thank you :)

You saved atleast 2 minutes by viewing the ad-free version of this page. Thank you for being a part of Teachoo Black.


Transcript

Given Height of pole = DE = 6 m Length of shadow = EF = 2βˆšπŸ‘m We need to find Angle of elevation of the sun i.e 𝛉 In βˆ† 𝐷𝐸𝐹 tan πœƒ = (𝑆𝑖𝑑𝑒 π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ π‘‘π‘œ π‘Žπ‘›π‘”π‘™π‘’" " 𝐹)/β–ˆ(𝑆𝑖𝑑𝑒 π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ π‘‘π‘œ π‘Žπ‘›π‘”π‘™π‘’" " 𝐹@ ) tan 𝜽 = (" " 𝑫𝑬)/𝑬𝑭 tan πœƒ = (" " 6)/(2√3) tan πœƒ = (" " 3)/√3 tan πœƒ =√3 ∴ 𝜽 = 60 Β° So, the correct answer is (a)

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo