Question

In Fig. 12.58, what are the angles of depression from the observing positions O1 and O2 of the object at A?

Answer 1

angle of depression from the observing positions O₁  = 30° & angle of depression from the observing positions O₂  =  45°

Step-by-step explanation:

See the attached  Figure

angle of depression from the observing positions O₁  = 90° - 60°

=> angle of depression from the observing positions O₁  = 30°

∠A = 45°

=> ∠AO₂B = 90⁰ - 45⁰ = 45⁰

=> ∠AO₂B = 45⁰

angle of depression from the observing positions O₂  = 90° - 45°

= 45°

angle of depression from the observing positions O₂  =  45°

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Answer 2

The angles of depression from the observing positions of $O_1$ and $O_2$ are 30° and 45°

Explanation:

Let x be the angle of depression of the object at A from point $O_1$

Let y be the angle of depression of the object at B from point $O_2$

Let us consider the triangle $Q_1CA$

By angle sum property, we have,

$\angle \mathrm{O}_{1} \mathrm{AC}+\angle \mathrm{ACO}_{1}+\angle \mathrm{AO}_{1} \mathrm{C}=180^{\circ}$

             $\angle \mathrm{O}_{1} \mathrm{AC}+90^{\circ}+60^{\circ}=180^{\circ}$

                     $\angle \mathrm{O}_{1} \mathrm{AC}+150^{\circ}=180^{\circ}$

                                 $\angle \mathrm{O}_{1} \mathrm{AC}=30^{\circ}$

Let us draw a line $\mathrm{O}_{1} \mathrm{M}$ parallel to AC

Also, let us draw a line $\mathrm{O}_{2} \mathrm{N}$ parallel to AC

It is obvious that $\mathrm{O}_{1} \mathrm{A}$ is a transversal line, then,

$\ {x}=\angle \mathrm{O}_{1} \mathrm{AC}$

$x=30^{\circ}$

Also, $\mathrm{O}_{2} \mathrm{A}$ is a transversal line, then,

$\ y=\angle \mathrm{O}_{2} \mathrm{AB}$

$y=45^{\circ}$

Thus, the angles of depression from the observing positions of $O_1$ and $O_2$ are 30° and 45°

Learn more:

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