Trigonometry

Question

AB is a hill. Points D and C are in line with the base B of the hill. From D, the angle of elevation of the top of the hill is 24º and from C, 40 metres closer to the hill, the elevation is 52º. Let AB = h metres and BC = x metres. 

i) Show that h=(x+40)tan24

ii) Hence, show that h=40tan52tan24/tan52-tan24

iii) Find the value of h, to 3 sig. figs.

Answer 1

 

 

i)

 

  From right triangle ADB we have 

   tan 24 = h/(x+40) ----> h = (x+40)*tan24

 

 

(ii)  From right triangle ABC we have:

              tan 52 = h/x ---> h = x * tan 52 ---> x = h/tan52

 

        substituiting into (i):   h = {( h/tan52) + 40} * tan 24

                                           =  tan 24*{h + 40*tan52}/tan52  <--- common denominator of tan 52

 

 So then h * tan 52 = tan24 * { h + 40*tan52}  <--- multiplies both sides by tan 52

 

             h * tan 52 = tan24*h + 40*Tan52*tan24 <--- distributive

 

             h * tan 52 - tan 24*h = 40*tan52*tan24 <--- subtracts tan24*h from both sides

 

           h ( tan 52 - tan 24) = 40 * tan 52 * tan 24  <--- factors out h from left side

 

             h = (40 * tan 52 * tan24)/(tan 52 - tan 24) <---- solves for h

 

 

(iii) 27.308 is the evaluation if (ii) to 3 significant figures