Question

~QUIZ~
.
Soal:
tan 75° = . . . ?
.
Syarat untuk menjawab soal :
● Dilarang jawaban berupa komentar spam atau asal²an.
● Dilarang copas jawaban dari google.
● Jawabannya harus disertai dengan penjelasan yang masuk akal.
● Gunakanlah kata-kata jawabanmu sendiri yang baik dan benar.

Answer 1

Jawaban:

2+√3

Penjelasan dengan langkah-langkah:

[tex] \tan \: 7 {5} = tan \: (45 + 30) \\ = \frac{tan \: 45 + tan \: 30}{1 - tan45.tan30} \\ = \frac{1 + \frac{1}{3} \sqrt{3} }{1 - 1. \frac{1}{3} \sqrt{3} } \\ = \frac{ \frac{3 + \sqrt{3} }{3} }{ \frac{3 - \sqrt{3} }{3} } \\ = \frac{3 + \sqrt{3} }{3 - \sqrt{3} } \\ rasionalkan \\ = \frac{3 + \sqrt{3} }{3 - \sqrt{3} } \times \frac{3 + \sqrt{3} }{3 + \sqrt{3} } \\ = \frac{9 + 6 \sqrt{3} + 3 }{9 - 3} \\ = \frac{12 + 6 \sqrt{3} }{6} \\ = 2 + \sqrt{3} [/tex]

CMIIW

Answer 2

Verified answer

[tex]\sf \bf \sf tan \: 75°[/tex]

[tex]\sf \bf \sf tan \: (45° + 30°)[/tex]

[tex]\sf \bf \sf \frac{tan \: 45° + tan \: 30°}{1- tan \: 45° ( tan \: 30° )}[/tex]

[tex]\sf \bf \sf \frac{1 + \frac{1}{3} \sqrt{3}}{1- 1 ( \frac{1}{3} \sqrt{3})}[/tex]

[tex]\sf \bf \sf \frac{1 + \frac{1}{3} \sqrt{3}}{1- \frac{1}{3} \sqrt{3}} × \frac{1+\frac{1}{3} \sqrt{3}}{1+ \frac{1}{3} \sqrt{3}}[/tex]

[tex]\sf \bf \sf \frac{1 + \frac{1}{3} \sqrt{3} + \frac{1}{3}\sqrt{3} + \frac{1}{3}}{1 - \frac{1}{3}}[/tex]

[tex]\sf \bf \sf \frac{\frac{1}{1}+ \frac{1}{3} \sqrt{3} + \frac{1+1}{3}(\sqrt{3})}{\frac{1}{1}- \frac{1}{3}}[/tex]

[tex]\sf \bf \sf \frac{\frac{1×3}{1×3}+ \frac{1}{3} \sqrt{3} + \frac{2}{3} \sqrt{3}}{\frac{1 × 3}{1 × 3}- \frac{1}{3}}[/tex]

[tex]\sf \bf \sf \frac{\frac{3}{3} + \frac{1}{3} \sqrt{3} + \frac{2}{3} \sqrt{3}}{\frac{3}{ 3}- \frac{1}{3}}[/tex]

[tex]\sf \bf \sf \frac{ \frac{3+1}{3} \sqrt{3} + \frac{2}{3} \sqrt{3}}{\frac{3 - 1}{ 3}}[/tex]

[tex]\sf \bf \sf \frac{\frac{4}{3} \sqrt{3} + \frac{2}{3} \sqrt{3}}{\frac{2}{ 3}}[/tex]

[tex]\sf \bf \sf ( \frac{4}{3} + \frac{2}{3} \sqrt{3} ) × \frac{3}{2}[/tex]

[tex]\sf \bf \sf \frac{4(3)}{3(2)} + \frac{2(3)}{3(2)} \sqrt{3} [/tex]

[tex]\sf \bf \sf \frac{12}{6} + \frac{6}{6} \sqrt{3} [/tex]

[tex]\sf \bf \sf 2 + 1 \sqrt{3} [/tex]

[tex]\bold{\underline{\boxed{\pink{\sf \bf \tt 2 + \sqrt{3}}}}}[/tex]

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