Trigonometry

Question

A man standing on a hill that is 500 ft high sees a bear walking straight toward the hill. He measures the angle of depression to the bear to be 10, and then measures again 5 minutes later to find the angle of depression to be 12. How far did the bear walk in 5 minutes?

Answer 1

Hi Zoe,

 

This problem is another variation of your earlier problem.

 

Click this link (https://dl.dropbox.com/s/n8x7po5cockb6jr/Bear_Approachinng_Hill.png?raw=1) to view our diagram for this problem.

 

The man is standing on top of a hill (point B) and looks down at an angle of depression of 10 degrees relative to a bear on the ground (point D) approaching the hill.  Five minutes later, the bear is closer to the hill (point C) with the man now looking down at an angle of depression of 12 degrees relative to the approaching bear.

 

The distance that the bear moved in 5 minutes is the length of CD, which is the length of BD minus the length BC.

 

In right triangle ABD, we have:

    tan 10 = AB / BD

    tan 10 = 500 / BD

    BD = 500 / tan 10

    BD = 2835.6 feet

 

In right triangle ABC, we have:
    tan 12 = AB / BC
    tan 12 = 500 / BC
    BC = 500 / tan 12
    BC = 2352.3 feet

 

Finally, subtracting length of BC from length of BD giving length of CD:

    CD = BD - BC

    CD = 2835.6 - 2352.3

    CD = 483.3 feet (distance bear moved in 5 minuts)

 

Once again, getting the right diagram was crucial to the solution.

 

Thanks for submitting another problem and glad to help again.

 

God bless, Jordan. 

 

Answer 2

 

Draw the figure for this first .make a right angled triangle with right angle at one end of the base and other end as angle of measure 10 degree.Mark each angle let base is B(90 degree) and C(10 degree) and the third vertex as A .As height of the hill is given as 500 ft ,mark AB as 500 ft.when bear moves toward the hill after 5 min. angle of depression chenges to 12 degree,so mark a point D on the base BC ,in between the points B and C now ∠ ADB=12 degree

let BD=x and DC = y  ==> BC=x+y

 

consider right angled triangle ABC,  

tan 10 = AB/BC

tan 10 = 500/(x+y)

 x+y = 500/tan 10 =2835.6 ft

 

 next take right angled triangle ABD,

tan 12 =AB/ BD

tan 12 =500/x

x= 500/tan12

  =2352.3 ft

 

therefore distance travelled by bear in 5 min = BC-BD  

                                                                  =(x+y)-x

                                                                  =2835.6-2352.3

                                                                  = 483.3ft