Question

fast
class 10
chapter 8or 9
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Answer 1

Given,

Angle of elevation at distance S +Angle of elevation at distance T=90°

Refer to diagram given in attachment,

∠ADB+∠ACB=90°

Let, ∠ADB=Ф

Then, ∠ACB=90°-Ф

Solution,

In ΔADB,

tanФ=AB/BD=H/T

In ΔACB,

tan(90-Ф)=CotФ                ( here Ф refers to ∠ADB )

=>H/S=T/H

Now,

tanФ=H/T......(i) and CotФ=H/S......(ii)

By multiplying i and ii,we get

tanФ.cotФ=H²/ST

=>1=H²/ST      (tanФ.cotФ=1)

=>H²=ST

=>H=√ST

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