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in Trigonometry by (47.6k points)
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A ladder on the platform of a fire brigade van can be elevated at an angle of 70° to the maximum. The length of the ladder can be extended upto 20 m. If the platform is 2 m above the ground, find the maximum height from the ground upto which the ladder can reach. (sin 70° = 0.94)

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Let AB represent the length of the ladder and AE represent the height of the platform.

Draw seg AC ⊥ seg BD. 

Angle of elevation = ∠BAC = 70° 

AB = 20m 

AE = 2m 

In right-angled ∆ABC,

sin 70° = BC/AB …..[By definition] 

∴ 0.94 = BC/20

∴ BC = 0.94 × 20 

= 18.80 m 

In \(\Box\)ACDE, 

∠E = ∠D = 90° 

∠C = 90° … [seg AC ⊥ seg BD] 

∴ ∠A = 90° … [Remaining angle of \(\Box\)ACDE] 

\(\Box\)ACDE is a rectangle. … [Each angle is 90°] 

∴ CD = AE = 2 m … [Opposite sides of a rectangle] 

Now, BD = BC + CD … [B – C – D] 

= 18.80 + 2 

= 20.80 m 

∴ The maximum height from the ground upto which the ladder can reach is 20.80 metrs.

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