WebThe shadow of a vertical tower on level ground increases by 20 m, when the altitude of the sun changes from angle of elevation 60 o to 45 o. Find the height of the tower. A 47.32 m B 47 m C 48 m D 49.56 m Medium Solution Verified by Toppr Correct option is A) Let the height of the tower be h In right ΔOBC tan60 0= xh =>3= xh WebMar 9, 2024 · The shadow of a tower standing on level ground is found to be 40 m longer when the sun's altitude is 30 ∘ than when it is 60 ∘. Find the height of the tower. Last updated date: 16th Jan 2024 • Total views: 185.1k • Views today: 4.83k Answer Verified 185.1k + views Hint: First draw a figure so that you can understand.
The shadow of a vertical tower on a level ground increases by
WebThe length of'the shadow of a vertical tower on level ground increases by 10 metres when the altitude of the sun changes from 45° to 30°. Then the height of the tower is A) 5 3 m e t r e B) 10 ( 3 + 1) m e t r e C) 5 ( 3 + 1) m e t r e D) 10 3 m e t r e Correct Answer: C) 5 ( 3 + 1) m e t r e Description for Correct answer: AB = tower = h WebApr 12, 2024 · UAV Towers. First seen in DMZ, UAV Towers can be activated during a match to provide intel on enemy positioning in the form of several circular UAV sweeps from the tower’s fixed location. ... Missions are now shown in a vertical list with new mission buttons. Shortened description panel so that selected Missions can be viewed on the right ... sunova koers
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WebApr 22, 2024 · Hi, You can visit our telegram channel to get the the notes of video.Telegram - Solutions with Mishra sirI know there are many channels who are already provi... WebThe shadow of a vertical tower on a level ground increases by 10 when the altitude of the sun changes from 45 0 to 30 0. Find the height of the tower, correct to two decimal … WebThe length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun's elevation is 30∘ than when it was 45∘. The height of the tower is (a) (2√3x) m (b) (3√2x) m (c) (√3−1)x m (d) (√3+1)x m Solution Class X in BCD, tan450 = h y → h= y in ABC, tan300 = h 2x+y 1 √3 = h 2x+y 2x =(√3−1)h (since h=y ) sunova nz