The upper ((3/4))th portion of a vertical pole subtends an angle (tan)- 1((3/5)) at a point in the horizontal plane through its foot and at a distance 40 m from the foot. The height of vertical pole is :-

Question

The upper ((3/4))th portion of a vertical pole subtends an angle (tan)- 1((3/5)) at a point in the horizontal plane through its foot and at a distance 40 m from the foot. The height of vertical pole is :- (A) 20 m (B) 40 m (C) 60 m (D) 80 m

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-hxvmoctmdwceumya.jpg" /> tan(θ + α )=(x/40) &tanθ =(x/160) tanα =(3/5) ∴ α =(.(.θ +α .)-θ .) ⇒ tanα =tan(.(.θ +α .)-θ .) ⇒ (3/5)=((x/40) - (x/160)/1 + (x)40 ⋅ (x)160 ∴ x=40 m

Q. The upper $(\frac{3}{4})^{t h}$ portion of a vertical pole subtends an angle $(t an)^{- 1} (\frac{3}{5})$ at a point in the horizontal plane through its foot and at a distance $40 m$ from the foot. The height of vertical pole is :-

$20 m$

$40 m$

$60 m$

$80 m$

$t an (θ + α) = \frac{x}{40}$
$& t an θ = \frac{x}{160}$
$t an α = \frac{3}{5}$
$∴ α = ((θ + α) - θ)$
$\Rightarrow t an α = t an ((θ + α) - θ)$
$\Rightarrow \frac{3}{5} = \frac{\frac{x}{40} - \frac{x}{160}}{1 + \frac{x}{40} \cdot \frac{x}{160}}$
$∴ x = 40 m$

Q. The upper (43​)th portion of a vertical pole subtends an angle (tan)−1(53​) at a point in the horizontal plane through its foot and at a distance 40m from the foot. The height of vertical pole is :-

20m

40m

60m

80m