(sin 150\xb0+cos 30\xba), Nilai dari, adalah .., (tan 225\xba-sin 300\xb0)

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(sin 150°+cos 30º)
Nilai dari
adalah ..
(tan 225º-sin 300°)​

~Math

sin 150°+cos 30º

sin 150° = sin ( 180 – 30 )

= sin 30

= 1/2

cos 30 = 1/2 √3

sin 150°+cos 30º = 1/2 + 1/2 √3

sin 150°+cos 30º = 1 + √3 /2

_____________________

tan 225º-sin 300°

tan 225° = tan ( 180 + 45 )

= tan 45

= 1

sin 300 = sin ( 360 – 60 )

= sin 60

= 1/2 √3

tan 225º-sin 300° = 1 – 1/2 √3

tan 225º-sin 300° = 2 – √3 /2

#balajar bersam brainly


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