(will give brainliest) a hiker, whose eye is 1.6 meters above ground, stands 50 meters from the base of connecting her eye and the top of the cliff and a horizontal line is 58°. draw a diagram representing the situation and find the height of the cliff

Step-by-step explanation:

Find the diagram attached. Using SOH, CAH, TOA to find the height of the cliff AC,

AC is the opposite side and BC is the adjacent

Given BC = 50m and ∠ABC = 58°

tan∠ABC = AC/BC

tan58° = AC/50

cross multiply

AC = 50tan58°

AC = 80.02m

Hence the height of the cliff is 80.02m

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.

cosine

step-by-step explanation: