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While flying at an altitude of 1.5 km, a plane measures angles or depression to opposite ends of a large crater, shown in the image below. Find the width of the crater

While flying at an altitude of 15 km a plane measures angles or depression to opposite ends of a large crater shown in the image below Find the width of the cra class=

Respuesta :

Check the picture below.

notice the alternate interior angles in the picture.

[tex]\bf tan(68^o)=\cfrac{\stackrel{opposite}{1.5}}{\stackrel{adjacent}{x}}\implies x=\cfrac{1.5}{tan(68^o)}\implies x\approx 0.61 \\\\[-0.35em] ~\dotfill\\\\ tan(56^o)=\cfrac{\stackrel{opposite}{1.5}}{\stackrel{adjacent}{w}}\implies w=\cfrac{1.5}{tan(56^o)}\implies w\approx 1.01 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{\textit{width of the crater}}{x+w\implies 1.62}~\hfill[/tex]

Ver imagen jdoe0001

The width of the crater is 1.61 km.

Given that:

Plane is flying at altitude of  1.5 km

The angle of depressions to point A = 68 degrees/

The angle of depression to point B = 56 degrees.

Calculations:

In the given diagram below, we can see:

Angle XPD is right angle and thus we have:

[tex]\angle APD + \angle APX = 90\\\angle APD = 180 - \angle APX = 90 - 68 = 22^\circ[/tex]

Similarly, Angle  YPD is right angle and thus:

[tex]\angle DPB + \angle BPY = 90\\\angle DPB = 90 - \angle BPY = 90 - 56 = 34^\circ[/tex]

Since the triangle ADP and triangle  PDB  are right angled triangles, thus we have by trigonometric ratios:

[tex]tan(22) = \dfrac{AD}{PD}\\\\0.404 \times 1.5 = AD\\\\AD = 0.606\: \rm km[/tex]

Similarly,

[tex]tan(34) = \dfrac{BD}{PD}\\\\0.674 \times 1.5 = BD\\\\BD = 1.01 \: \rm km[/tex]

The width of the crater = AD + DB = 1.01 km + 0.606 km  = 1.61 km

Thus, width of the crater is 1.61 km.

Learn more about angle of depression here:

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Ver imagen astha8579