Respuesta :

budwilkins
sin(θ) = cos(48°)
sin(θ) = 0.6691
θ = arcsin (0.6691)
θ = 42°

** for these types of problems, get familiar with the inverse functions on your calculator:
[tex]arcsin = { \sin}^{ - 1} \\ arccos = { \cos}^{ - 1} \\ arctan = { \tan}^{ - 1} [/tex]
JeanaShupp

Answer: [tex]\theta= 42^{\circ}[/tex]

Step-by-step explanation:

Given: For the acute angle [tex]\theta[/tex] in a right triangle.

[tex]\sin(\theta)=\cos(48^{\circ})[/tex]............................................(1)

We know that [tex]\cos\theta=\sin(90^{\circ}-\theta)[/tex]

Then by using the above identity we have

[tex]\cos(48^{\circ})=\sin(90^{\circ}-48^{\circ})=\sin(42^{\circ})[/tex]............................(2)

From (1) and (2), we get

[tex]\sin(\theta)=\sin(42^{\circ})\\\\\Rightarrow\theta= 42^{\circ}[/tex]