Question

A vertical tower stands on a horizontal plane and is
surmounted by a vertical flag staff of height h. At a
point on the plane the angle of elevation of the bottom
of the flag staff of a and that of the top of the flag staff
is ß. Then the height of the tower is
$(1) \: h \: tan \alpha$
$(2) \frac{htan \alpha }{tan \beta - tan \alpha }$
$(3) \: \frac{h \: tan \alpha }{h \: tan \alpha - h \: tan \beta }$
$(4) \: none \: of \: these$






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Answer 1

ANSWER

Let height be y ΔOAC

Let height be y ΔOACtanθ=BP

Let height be y ΔOACtanθ=BPtanβ=OACA

Let height be y ΔOACtanθ=BPtanβ=OACAtanβ=xy+h            (y+h)→ Let AB, AB+BC

Let height be y ΔOACtanθ=BPtanβ=OACAtanβ=xy+h            (y+h)→ Let AB, AB+BCLet OA→ x

Let height be y ΔOACtanθ=BPtanβ=OACAtanβ=xy+h            (y+h)→ Let AB, AB+BCLet OA→ xx=[tanβy+x]

Let height be y ΔOACtanθ=BPtanβ=OACAtanβ=xy+h            (y+h)→ Let AB, AB+BCLet OA→ xx=[tanβy+x]Consider ΔOAB

Let height be y ΔOACtanθ=BPtanβ=OACAtanβ=xy+h            (y+h)→ Let AB, AB+BCLet OA→ xx=[tanβy+x]Consider ΔOABtanα=xy=Baseperpendicular

Let height be y ΔOACtanθ=BPtanβ=OACAtanβ=xy+h            (y+h)→ Let AB, AB+BCLet OA→ xx=[tanβy+x]Consider ΔOABtanα=xy=Baseperpendicularx=tanαy

Let height be y ΔOACtanθ=BPtanβ=OACAtanβ=xy+h            (y+h)→ Let AB, AB+BCLet OA→ xx=[tanβy+x]Consider ΔOABtanα=xy=Baseperpendicularx=tanαytanαy=tanβy+h

Let height be y ΔOACtanθ=BPtanβ=OACAtanβ=xy+h            (y+h)→ Let AB, AB+BCLet OA→ xx=[tanβy+x]Consider ΔOABtanα=xy=Baseperpendicularx=tanαytanαy=tanβy+hytanβ=tanαy+tanαh

Let height be y ΔOACtanθ=BPtanβ=OACAtanβ=xy+h            (y+h)→ Let AB, AB+BCLet OA→ xx=[tanβy+x]Consider ΔOABtanα=xy=Baseperpendicularx=tanαytanαy=tanβy+hytanβ=tanαy+tanαhytanβ−tany=tanαh

Let height be y ΔOACtanθ=BPtanβ=OACAtanβ=xy+h            (y+h)→ Let AB, AB+BCLet OA→ xx=[tanβy+x]Consider ΔOABtanα=xy=Baseperpendicularx=tanαytanαy=tanβy+hytanβ=tanαy+tanαhytanβ−tany=tanαhy(tanβ−tanα)=tanαh

Let height be y ΔOACtanθ=BPtanβ=OACAtanβ=xy+h            (y+h)→ Let AB, AB+BCLet OA→ xx=[tanβy+x]Consider ΔOABtanα=xy=Baseperpendicularx=tanαytanαy=tanβy+hytanβ=tanαy+tanαhytanβ−tany=tanαhy(tanβ−tanα)=tanαhy=tanβ−tanαhtanα

Let height be y ΔOACtanθ=BPtanβ=OACAtanβ=xy+h            (y+h)→ Let AB, AB+BCLet OA→ xx=[tanβy+x]Consider ΔOABtanα=xy=Baseperpendicularx=tanαytanαy=tanβy+hytanβ=tanαy+tanαhytanβ−tany=tanαhy(tanβ−tanα)=tanαhy=tanβ−tanαhtanαThis proved.

Answer 2

(2) is correct option