If,
[tex] \tan \: theta \cos60 = \frac{ \sqrt{3} }{2} [/tex]
then,
[tex] \sin(theta - 15) = [/tex]
(a)
[tex] \frac{1}{ \sqrt{2} } [/tex]
(b) 1
(c)
[tex] \sqrt{2} [/tex]
(d) 0
[tex] \tan \: theta \cos60 = \frac{ \sqrt{3} }{2} [/tex]
then,
[tex] \sin(theta - 15) = [/tex]
(a)
[tex] \frac{1}{ \sqrt{2} } [/tex]
(b) 1
(c)
[tex] \sqrt{2} [/tex]
(d) 0
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Answer:
If,
tan
�
ℎ
�
�
�
cos
60
=
3
2
tanthetacos60=
2
3
then,
sin
(
�
ℎ
�
�
�
−
15
)
=
sin(theta−15)=
(a)
1
2
2
1
(b) 1
(c)
2
2
(d) 0