A vertical pole PS has two marks at Qand A such that the portions PQ- PR and PS subtend angles - - at a point on the ground distant x from the pole- If PQ - a- PR -b- PS - c and - - - - - - 180- then x2 -

The correct option is D abca + b + c PQ = a = x tan α , PR = b = x tan β PS = c = tab γ a + b + c = x ( tan α + tan β + tan ...

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Standard X

Mathematics

Calculating Heights and Distances

A vertical po...

Question

A vertical pole PS has two marks at Qand A such that the portions PQ, PR and PS subtend angles $α, β, γ$ at a point on the ground distant x from the pole. If PQ = a, PR -b, PS = c and $α + β + γ$ = 180; then $x^{2}$ =

\frac{a}{a + b + c}

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\frac{b}{a + b + c}

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\frac{c}{a + b + c}

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\frac{a b c}{a + b + c}

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Solution

The correct option is D $\frac{a b c}{a + b + c}$
PQ = a = x tan $α$ , PR = b = x tan $β$
PS = c = tab $γ$
a + b + c = x ( tan $α$ + tan $β$ + tan $γ$ )
abc = $x^{3}$ .tan $α$ .tan $β$ . tan $γ$
Dividing abc/a + b +c = $x^{2}$ .1 $\Rightarrow$ (d)
when $α$ + $β$ + $γ$ = $180^{0}$ then
tan $α$ + tan $β$ + tan $γ$ = tan $α$ .tan $β$ . tan $γ$
i.e. $S_{1} = S_{3}$

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A vertical pole PS has two marks at Qand A such that the portions PQ, PR and PS subtend angles α,β,γ at a point on the ground distant x from the pole. If PQ = a, PR -b, PS = c and α+β+γ = 180; then x2 =

The correct option is D abca+b+cPQ = a = x tan α, PR = b = x tan βPS = c = tab γa + b + c = x ( tan α + tan β + tan γ)abc = x3.tan α.tan β. tan γDividing abc/a + b +c = x2.1 ⇒ (d)when α + β + γ = 1800 thentan α + tan β + tan γ = tan α.tan β. tan γi.e. S1=S3

A vertical pole PS has two marks at Qand A such that the portions PQ, PR and PS subtend angles $α, β, γ$ at a point on the ground distant x from the pole. If PQ = a, PR -b, PS = c and $α + β + γ$ = 180; then $x^{2}$ =