Example 1 - A tower stands vertically on the ground. From - Questions easy to difficult

Example 1 - Chapter 9 Class 10 Some Applications of Trigonometry - Part 2


Transcript

Example 1 A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60Β°. Find the height of the tower. Let, the height of the tower be H meter. So, AB = H metre Distance of the point from the foot of the tower = 15m Hence, CB = 15m Angle of elevation = 60Β° ∠ACB = 60Β° Since tower is vertical to ground, So, ∠ ABC = 90Β° Now, tan C = (𝑆𝑖𝑑𝑒 π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ π‘‘π‘œ π‘Žπ‘›π‘”π‘™π‘’" c " )/(𝑆𝑖𝑑𝑒 π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ π‘‘π‘œ π‘Žπ‘›π‘”π‘™π‘’" " 𝐢) tan C = (" " 𝐴𝐡)/𝐢𝐡 tan 60Β° = 𝐴𝐡/𝐢𝐡 √3 = 𝐻/15 15√3 = H H = 15√3 Hence, height of the tower = H = 15√3 m

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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, Social Science, Physics, Chemistry, Computer Science at Teachoo.