A ski jumper is preparing for her start on top of the jumping hill in korea which has a horizontal run of 234 meters. She sees the end of the hill under a 36 degree angle of depression. How high is the jumping hill?

The height of the jumping hill x = 189.3 m

Step-by-step explanation:

From Δ ABC

AB = x = height of the hill

∠A = 36° , ∠B = 90°

Thus ∠C = 180 - ∠A - ∠B

⇒ ∠C = 180 - 36 - 90

⇒ ∠C =  54°

From Δ ABC  

x = 0.809 × 234

x = 189.3 m

This is the height of the jumping hill.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.