Answer:
[tex]\bold{sin(38^\circ)=cos(52^\circ)}[/tex]
Step-by-step explanation:
Given that [tex]\triangle KJL[/tex] is a right angled triangle.
[tex]\angle JKL = 52^\circ\\\angle KLJ = 38^\circ[/tex]
and
[tex]\angle KJL = 90^\circ[/tex]
Kindly refer to the attached image of [tex]\triangle KJL[/tex] in which all the given angles are shown.
To find:
sin(38°) = ?
a) cos(38°)
b) cos(52°)
c) tan(38°)
d) tan(52°)
Solution:
Let us use the trigonometric identities in the given [tex]\triangle KJL[/tex].
We have to find the value of sin(38°).
We know that sine trigonometric identity is given as:
[tex]sin\theta =\dfrac{Perpendicular}{Hypotenuse}[/tex]
[tex]sin(\angle JLK) = \dfrac{JK}{KL}\\OR\\sin(38^\circ) = \dfrac{JK}{KL}[/tex]....... (1)
Now, let us find out the values of trigonometric functions given in options one by one:
[tex]cos\theta =\dfrac{Base}{Hypotenuse}[/tex]
[tex]cos(\angle JLK) = \dfrac{JL}{KL}\\OR\\cos(38^\circ) = \dfrac{JL}{KL}[/tex]....... (2)
By (1) and (2):
sin(38°) [tex]\neq[/tex] cos(38°).
[tex]cos(\angle JKL) = \dfrac{JK}{KL}\\OR\\cos(52^\circ) = \dfrac{JK}{KL}[/tex] ...... (3)
Comparing equations (1) and (3):
we get the both are same.
[tex]\therefore \bold{sin(38^\circ)=cos(52^\circ)}[/tex]
