Trigonometry

Question

from the top of a building a person spots two boats out on the lake. the angle of depression from the person to one of the boats is 48 degrees, the angle of depression from the person to the other boat is 20 degrees. how far apart are the boats if the person is 10 meters above the ground

Answer 1

Distance of the building from the Boat at 20 degree angle of depression = 10/tan 20º = 0.36397023426620236135104788277683 meters

Distance of the building from the Boat at 48 degree angle of depression = 10/tan 48º = 1.1106125148291928701434819641651 meters

 

Therefore, distance between the boats= 0.7466422805629905087924340813883 meters 

 

Assuming the base of the buiding on top of which the man is standing and the boats are at the same elevation on the earth.

Answer 2

Draw a right triangle.  Label the right angle vertex A, vertex B is the top of the triangle and vertex C is to the right of A.  Place a point D between A and C.  D is the boat with the angle of depression of 48 degrees and C is the boat with an angle of depression ot 20 degrees.  Draw a line segment from B to D.  Draw a horizontal line through B and place a point on the line to the right of B.  Side AB is 10m, angle EBD is 48 degrees and angle EBC is 20 degrees.  Label distances AD x and DC y.  Using the drawing find angle ABD.  ABD = 90 - 48 = 52 degrees and ABC = 90 - 20 = 70 degrees.  Use the trig function tangent to set up the equations.

 

eq1:  tan(52°) = x/10

 

eq2:  tan(70°) = (x+y)/10

 

Solve the first equation for x.  Substitute that x-value into the second equation for y.  If you cannot solve for y just let me know.

Answer 3

Draw and label a diagram.

The base of the building is C

The top of the building is D

The farthest boat (20° depression) is A

The closer boat is B

 

m∠DAB = 20°

tan 20° = DC / AC

tan 20° = 10 / AC

AC = 10 / tan 20°

AC ≈ 10 / 0.36397

AC ≈ 27.47

 

Calculate BC in the same way. m∠DBC = 48°

 

Distance between boats is AC minus BC.