Aradar operator on a ship discovers a large sunken vessel lying parallel to the ocean surface, 150 m
directly below the ship. the length of the vessel is a clue to which wreck has been found. the radar
operator measures the angles of depression to the front and back of the sunken vessel to be 52° and 70°.
how long, to the nearest tenth of a metre, is the sunken vessel?

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.

add an !

step-by-step explanation: