Question

6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of
elevation from his eyes to the top of the building increases from 30° to 60° as he walks
towards the building. Find the distance फ्रॉम ए पॉइंट ऑन द ग्राउंड द एंगल ऑफ एलिवेशन ऑफ द मोटर एंड द टॉप ऑफ ट्रांसमिशन टावर वर्क एट द टॉप ऑफ 20 मीटर हाई बिल्डिंग और 45 डिग्री एंड 60 डिग्री रिस्पेक्टिवली फाइंड द हाइट ऑफ द टावर ​

Answer 1

Let the boy initially standing at point Y with inclination 30° and then he approaches the building to the point X with inclination 60°.

∴ XY is the distance he walked towards the building. also, XY = CD.

Height of the building = AZ = 30 m AB = AZ - BZ = (30 - 1.5) = 28.5 m A/q

In right ΔABD, tan 30° = AB/BD ⇒ 1/√3 = 28.5/BD ⇒ BD = 28.5√3 m also , In right ΔABC, tan 60° = AB/BC ⇒ √3 = 28.5/BC ⇒ BC = 28.5/√3 = 28.5√3/3 m

∴ XY = CD = BD - BC = (28.5√3 - 28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 m

. Thus, the distance boy walked towards the building is 57/√3 m.

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