Respuesta :
Answer:
Therefore,
The correct options are
[tex]\sin x = \dfrac{c}{b}\\\\x=sin^{-1}(\dfrac{c}{b})[/tex]
[tex]\tan x = \dfrac{c}{a}\\\\x=tan^{-1}(\dfrac{c}{a})[/tex]
[tex]\cos y = \dfrac{c}{b}\\\\x=cos^{-1}(\dfrac{c}{b})[/tex]
Step-by-step explanation:
Consider a right angle Triangle ΔABC, such that
AB = c = Adjacent side opposite to angle x
BC = a = Adjacent side opposite to angle y
AC = b = Hypotenuse
To Check :
For correct option?
Solution:
In Right Angle Triangle ABC,
[tex]\sin x = \dfrac{\textrm{side opposite to angle x}}{Hypotenuse}\\[/tex]
[tex]\cos x = \dfrac{\textrm{Adjacent side to angle x}}{Hypotenuse}\\[/tex]
[tex]\tan x = \dfrac{\textrm{side opposite to angle x}}{\textrm{side adjacent to angle x}}[/tex]
Substituting the values we get
[tex]\sin x = \dfrac{c}{b}\\\\x=sin^{-1}(\dfrac{c}{b})[/tex]
[tex]\cos x = \dfrac{a}{b}\\\\x=cos^{-1}(\dfrac{a}{b})[/tex]
[tex]\tan x = \dfrac{c}{a}\\\\x=tan^{-1}(\dfrac{c}{a})[/tex]
Now for y
[tex]\sin y = \dfrac{a}{b}\\\\x=sin^{-1}(\dfrac{a}{b})[/tex]
[tex]\cos y = \dfrac{c}{b}\\\\x=cos^{-1}(\dfrac{c}{b})[/tex]
Therefore,
The correct options are
[tex]\sin x = \dfrac{c}{b}\\\\x=sin^{-1}(\dfrac{c}{b})[/tex]
[tex]\tan x = \dfrac{c}{a}\\\\x=tan^{-1}(\dfrac{c}{a})[/tex]
[tex]\cos y = \dfrac{c}{b}\\\\x=cos^{-1}(\dfrac{c}{b})[/tex]
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