Question

HELP ASAP! What is the value of θ for the acute angle in a right triangle?

sin(θ)=cos(48°)



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Answer 1

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:

`arcsin = { \sin}^( - 1) \\ arccos = { \cos}^( - 1) \\ arctan = { \tan}^( - 1)`

Answer 2

Answer:
`\theta= 42^(\circ)`

Explanation:

Given: For the acute angle
`\theta` in a right triangle.

`\sin(\theta)=\cos(48^(\circ))`............................................(1)

We know that
`\cos\theta=\sin(90^(\circ)-\theta)`

Then by using the above identity we have

`\cos(48^(\circ))=\sin(90^(\circ)-48^(\circ))=\sin(42^(\circ))`............................(2)

From (1) and (2), we get

`\sin(\theta)=\sin(42^(\circ))\\\\\Rightarrow\theta= 42^(\circ)`