Question 10 (Choice 2) - CBSE Class 10 Sample Paper for 2022 Boards - Maths Standard [Term 2] - Solutions of Sample Papers for Class 10 Boards

Last updated at May 29, 2023 by Teachoo

A boy 1.7 m tall is standing on a horizontal ground, 50 m away from a building. The angle of elevation of the top of the building from his eye is 60°. Calculate the height of the building. (Take √3 = 1.73)

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Question 10 (Choice 2) A boy 1.7 m tall is standing on a horizontal ground, 50 m away from a building. The angle of elevation of the top of the building from his eye is 60°. Calculate the height of the building. (Take √3 = 1.73) Given that
boy is standing 50 m away from the building
∴ QB = 50 m
And, Boy is 1.7 m tall
∴ PQ = 1.7 m
Given that,
Angle of elevation of top of building from his eye = 60°
Hence, ∠APC = 60°
Now,
PC & QB are parallel lines
So, PC = QB = 50 m
And,
PQ and CB are parallel lines
So, PQ = CB = 1.7 m
We need to find height of the building
i.e. AB
In right angle triangle APC,
tan P = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑃)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑃)
tan P = AC/PC
tan 60° = 𝐀𝐂/𝟓𝟎
√3 = (AB − BC)/50
50√3 = AB − BC
50√3 = AB − 1.7
50√3 + 1.7 = AB
AB = 50√𝟑 + 1.7
AB = 50 × 1.73 + 1.7
AB = 86.5 + 1.7
AB = 88.2 m
∴ Height of building is 88.2 m

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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