The two palm trees are of equal heights and are standing opposite each other on either side of the river, which is 80 m wide. From a point O between them on the river the angles of elevation of the top of the trees are 60° and 30°, respectively. Find the height of the trees and the distances of the point O from the trees.

The two palm trees are of equal heights and are standing opposite

Question 34 (Choice - 1) - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Part 2
Question 34 (Choice - 1) - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Part 3
Question 34 (Choice - 1) - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Part 4
Question 34 (Choice - 1) - CBSE Class 10 Sample Paper for 2021 Boards - Maths Standard - Part 5

 

 

Note : This is similar to Ex 9.1, 10,  Chapter 9 Class 10 Applications of Trigonometry (NCERT Book)

Check the answer here https:// www.teachoo.com /1810/535/Ex-9.1--10---Two-poles-of-equal-heights-are-standing-opposite/category/Ex-9.1/

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Question 34 (Choice - 1) The two palm trees are of equal heights and are standing opposite each other on either side of the river, which is 80 m wide. From a point O between them on the river the angles of elevation of the top of the trees are 60° and 30°, respectively. Find the height of the trees and the distances of the point O from the trees.Let two trees be AB & CD So, Length of trees = AB = CD Also, Width of the river = 80 m So, BC = 80m Let point O be point between the trees We need to find height of trees i.e. AB & CD and Distance of the point from trees, i.e. BO & CO Since poles are perpendicular to ground ∠ ABP = 90° & ∠ DCP = 90° In right angle triangle ABO tan O = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑂)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑂) tan OP = (" " 𝐴𝐵)/𝑂𝐵 tan 60° = (" " 𝐴𝐵)/𝑂𝐵 √3 OB = AB AB = √3 OB CD = √3 OB In a right angle triangle DCO, tan O = (𝑆𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑂)/(𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒" " 𝑂) tan 30° = (" " 𝐶𝐷)/𝑂𝐶 (" " 1)/√3 = 𝐶𝐷/𝑂𝐶 𝑂𝐶/√3 = CD CD = 𝑂𝐶/√3 From (1) & (2) 𝑂𝐶/√3 = √3 OB OC = √3 ×√3 OB OC = 3OB Now, BC = BO + CO 80 = BO + CO 80 = OB + 3OB 80 = 4OB 4OB = 80 OB = 80/4 OB = 20 m Now, OC = BC – OB OC = 80 – 20 OC = 60 m From (2) CD = √3OB CD = √3 × 20 CD = 20√𝟑 m Hence, Length of the tree = CD = 𝟐𝟎√𝟑 m And Distance of point O from tree CD = CO = 60 m Distance of point O from tree AB = BO = 20 m

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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