Question #142874

At an angle of elevation 610 captain on a ship noticed a person on the top of a cliff waiving towards the ship. To see him better, captain moved his ship 92m closer to the cliff. The angle of elevation at that moment was 690 . Determine the height of the cliff to the nearest tenth of a meter.


1
Expert's answer
2020-11-08T18:05:21-0500

SOLUTION

Here is the sketch of the scenario;


First, we separate the triangles;



Here, we apply the Sine Rule;

asinA=bsinB=csinC\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}


In this case we apply a and c. i.e.


asinA=csinC\frac{a}{sinA}=\frac{c}{sinC}


92sin(8°)=csin(61°)\frac{92}{sin(8°)}=\frac{c}{sin(61°)}


c=92×sin(61°)sin(8°)c = \frac{92×sin(61°)}{sin(8°)}


c=578.16mc = 578.16m


Therefore, X=578.16mX = 578.16m


Then, considering the last triangle, which is right-angle triangle, we have;


We consider the SOHCAHTOA concept, where we apply the sine formula (SOH)

sin(69°)=H578.16sin(69°)=\frac{H}{578.16}


H=578.16×sin(69°)H= 578.16 × sin (69°)


H=539.76mH = 539.76m (To the nearest tenths)

H=539.8mH = 539.8m

Therefore, Height of the cliff is 539.8m539.8m


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