Скачать или смотреть Class : 10th | Chapter : 9 | Exercise : 9.1 | Question : 12 | HBSE | CBSE | NCERT | 2023-24

Class : 10th | Chapter : 9 | Exercise : 9.1 | Question : 12 | HBSE | CBSE | NCERT | 2023-24

Ex 9.1 Class 10 Maths
Question 12.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution:

Let AB be the building of height 7 m and EC be the height of the tower.

A is the point from where elevation of tower is 60° and the angle of depression of its foot is 45°.

EC = DE + CD

Also, CD = AB = 7 m. and BC = AD

To Find: EC = Height of the tower

Design a figure based on given instructions:



In right ΔABC,

tan 45° = AB/BC

1= 7/BC

BC = 7

Since BC = AD

So AD = 7

Again, from right triangle ADE,

tan 60° = DE/AD

√3 = DE/7

⇒ DE = 7√3 m

Now: EC = DE + CD

= (7√3 + 7) = 7(√3+1)

Therefore, height of the tower is 7(√3+1) m. Answer!

12. एक 7 मीटर ऊँचे भवन के शिखर से एक केबल टावर के शिखर का उन्नयन कोण 60° है और इसके पाद का अवनमन कोण 45° है। मीनार की ऊँचाई ज्ञात कीजिए।

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