Dave is standing 250 m from the building and measure the angle of elevation to the top of the building to be 39° if David eyes 158 cm from the ground what is the height of the building

The height of the building is H = 360.44 m

Step-by-step explanation:

From the Δ ABC

AB = x

CD = BE = 250 m

BC = DE = 158 m

Height of the building H = AB + BC = x + 158 ------ (1)

In the Δ ABC

x = 202.44 m

Put this value of x in equation (1) we get

Height of the building H = 202.44 + 158

H = 360.44 m

Therefore the height of the building is H = 360.44 m

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.