Please helpppp need pic

Question

Khrystyna and Chris are standing 300 feet away from each other on flat ground looking up at a crane. If Khrystyna’s angle of elevation to the crane is 49˚ and Chris’s is 78˚, what is the height of the crane to the nearest tenth of a foot?

Answer 1

desmos.com/calculator/rcjz3qas6z

Answer 2

If x is horizontal distance from Khrystina to bottom of craine is, then Chris,s distance is (x - 300). If h is height of crane,then x = h/tan49° and x - 300 = h/tan78°;

From here we get equation for h: h/tan49° - 300 = h/tan78° and h = 300tan49°·tan78°/(tan78° - tan49°) ≈ 450.8 foot

Answer 3

When the crane is in front of Chris and Khrystyna

AB = crane's height

BC = Distance from Chris to the crane

BK = Distance from Khrystyna to the crane

Since Chris and Khrystyna are 300ft apart

so we can restate BK into  

BK = BC + 300

From Chris point of view

Tan 78 = AB/BC

Tan 78 = AB/BC

      AB = BC * tan 78

From Khrystyna point of view

tan 49 = AB / BK

tan 49 = AB / (BC + 300)

     AB = tan 49 * (BC + 300)

              AB = AB

BC * tan 78 = tan 49 (BC + 300)

BC * tan 78 = BC* tan 49 + 300 * tan 49

BC * tan 78 - BC* tan 49 =  300 * tan 49

    3.554 BC = 345.110

BC = 97.097 feet

AB = BC * tan 78

      = 97.097 * tan 78

      = 456.8 feet

When the crane is in between Chris and Khrystyna

AB = crane's height

BC = Distance from Chris to the crane

BK = Distance from Khrystyna to the crane

Since Chris and Khrystyna are 300ft apart

so we can restate BK into  

BK = 300 - BC

From Chris point of view

Tan 78 = AB/BC

Tan 78 = AB/BC

      AB = BC * tan 78

From Khrystyna point of view

tan 49 = AB / BK

tan 49 = AB / (300- BC)

     AB = tan 49 * (300- BC)

AB = AB

 BC * tan 78 = tan 49 * (300- BC)

BC * tan 78 = 300 * tan 49 – BC * tan 49

BC * tan 78 + BC* tan 49 =  300 * tan 49

    5.815 BC = 345.110

BC = 59.346 feet

AB = BC * tan 78

      = 59.346 * tan 78

      = 279.2 feet