Question

Given isosceles right triangle CDE with m∠D = 90, sin C = sin D

Answer 1

Answer:True

Step-by-step explanation:

Given

$\angle D=90^{\circ}$

For isosceles triangle , CD=DE (x=y)

from figure,

$\sin C=\dfrac{y}{H}\\\\\sin D=\dfrac{x}{H}\\\\\text{CD=DE}\\\\\therefore \sin C=\sin D=\dfrac{x}{H}=\dfrac{y}{H}$

Thus, the given statement is true