Cos 80 Derajat

Cos 80 Derajat.

Blog Koma

– Setelah mempelajari materi “rumus jumlah dan selisih sudut pada trigonometri” dan materi “rumus hasil kali antara dua bentuk trigonometri” serta rumus trigonometri yang lainnya, pada artikel ini kita akan coba membahas tentang
Penerapan Rumus Trigonometri pada Soal-soal Bagian 1. Soal-soal yang melibatkan rumus-rumus trigonometri ini biasanya kita jumpai pada soal UJian Nasional, soal seleksi masuk perguruan tinggi baik negeri maupun swasta seperti SBMPTN, UM UGM, SIMAK UI, dan lain-lainnya. Hal mendasar yang harus kita perhatikan adalah ketelitian baik dalam menggunakan rumusnya atau dalam melakukan penjabaran dan perhitungannya. Langsung saja kita pelajari beberapa contoh soal berikut ini.

1). Tentukan nilai dari bentuk

$ \sin 20^\circ \sin xl^\circ \sin 80^\circ $

?

Penyelesaian :

Ada tiga cara yang akan kita sajikan dalam menyelesaikan soal nomor one :

Cara I :

*). Rumus Dasar yang kita gunakan adalah rumus perkalian fungsi trigonometri :

$ \sin A . \sin B = -\frac{i}{2} [ \cos (A+B) – \cos (A-B)] $

$ \sin A . \cos B = \frac{ane}{2} [ \sin (A+B) + \sin (A-B)] $

$ \sin ( 180^\circ – A) = \sin A $

$ \sin 100^\circ = \sin ( 180^\circ – 80^\circ ) = \sin 80^\circ $

*). Menyelesaikan soal :

$ \begin{align} \sin 20^\circ \sin xl^\circ \sin eighty^\circ & = (\sin twenty^\circ . \sin xl^\circ ) . \sin 80^\circ \\ & = (\sin 40^\circ . \sin xx^\circ ) . \sin 80^\circ \\ & = \left(-\frac{1}{2} [ \cos (40^\circ + 20^\circ) – \cos (twoscore^\circ – 20^\circ)] \correct) . \sin 80^\circ \\ & = \left(-\frac{1}{2} [ \cos lx^\circ – \cos 20^\circ ] \right) . \sin 80^\circ \\ & = \left(-\frac{1}{2} [ \frac{ane}{2} – \cos 20^\circ ] \right) . \sin 80^\circ \\ & = \left( – \frac{one}{4} + \frac{1}{2} \cos 20^\circ \correct) . \sin 80^\circ \\ & = – \frac{1}{four}\sin 80^\circ + \frac{1}{2} \sin 80^\circ . \cos 20^\circ \\ & = – \frac{1}{iv}\sin 80^\circ + \frac{1}{ii} (\sin 80^\circ . \cos twenty^\circ ) \\ & = – \frac{i}{4}\sin lxxx^\circ + \frac{1}{ii} \times \frac{1}{2} [ \sin (80^\circ + twenty^\circ ) + \sin (80^\circ – 20^\circ )] \\ & = – \frac{1}{4}\sin 80^\circ + \frac{1}{4} [ \sin 100^\circ + \sin 60^\circ ] \\ & = – \frac{1}{4}\sin 80^\circ + \frac{1}{4} [ \sin 80^\circ + \sin 60^\circ ] \\ & = – \frac{1}{four}\sin 80^\circ + \frac{1}{4} \sin eighty^\circ + \frac{one}{4} \sin 60^\circ \\ & = \frac{ane}{four} \sin threescore^\circ \\ & = \frac{1}{iv} \times \frac{ane}{2} \sqrt{3} \\ & = \frac{1}{eight} \sqrt{3} \end{marshal} $

jadi, nilai $ \sin twenty^\circ \sin 40^\circ \sin 80^\circ = \frac{i}{eight} \sqrt{3} . \, \heartsuit $.

Cara II :

*). Rumus Dasar yang kita gunakan adalah rumus perkalian fungsi trigonometri :

$ \sin A . \sin B = -\frac{1}{two} [ \cos (A+B) – \cos (A-B)] $

$ \cos A . \sin B = \frac{1}{2} [ \sin (A+B) – \sin (A-B)] $

*). Menyelesaikan soal :

$ \begin{align} \sin xx^\circ \sin xl^\circ \sin 80^\circ & = \sin 20^\circ . (\sin 40^\circ . \sin 80^\circ ) \\ & = \sin 20^\circ . (\sin lxxx^\circ . \sin twoscore^\circ ) \\ & = \sin twenty^\circ . \left( -\frac{1}{2} [ \cos (80^\circ + forty^\circ ) – \cos (80^\circ – twoscore^\circ )]\right) \\ & = \sin 20^\circ . \left( -\frac{i}{two} [ \cos 120^\circ – \cos forty^\circ ]\correct) \\ & = \sin xx^\circ . \left( -\frac{1}{2} [ -\frac{i}{2} – \cos xl^\circ ]\right) \\ & = \sin 20^\circ . \left( \frac{1}{4} + \frac{one}{2} \cos xl^\circ \right) \\ & = \frac{i}{iv} \sin xx^\circ + \frac{1}{two} \cos 40^\circ \sin 20^\circ \\ & = \frac{one}{iv} \sin 20^\circ + \frac{1}{ii} ( \cos 40^\circ \sin 20^\circ ) \\ & = \frac{1}{4} \sin 20^\circ + \frac{1}{two} \times ( \frac{1}{2} [ \sin (40^\circ + 20^\circ ) – \sin ( xl^\circ – 20^\circ )] ) \\ & = \frac{1}{iv} \sin 20^\circ + ( \frac{1}{4} [ \sin 60^\circ – \sin 20^\circ ] ) \\ & = \frac{one}{4} \sin 20^\circ + ( \frac{1}{4} [ \frac{1}{2}\sqrt{three} – \sin 20^\circ ] ) \\ & = \frac{i}{4} \sin 20^\circ + \frac{1}{8} \sqrt{3} – \frac{1}{iv} \sin 20^\circ \\ & = \frac{ane}{8} \sqrt{iii} \terminate{align} $

jadi, nilai $ \sin twenty^\circ \sin 40^\circ \sin lxxx^\circ = \frac{one}{viii} \sqrt{three} . \, \heartsuit $.

Baca :   Materi Fisika Kelas X Semester 2

Cara III :

*). Rumus Dasar yang kita gunakan adalah rumus perkalian fungsi trigonometri :

$ \sin A . \sin B = -\frac{1}{2} [ \cos (A+B) – \cos (A-B)] $

$ \cos A . \sin B = \frac{1}{2} [ \sin (A+B) – \sin (A-B)] $

$ \sin ( 180^\circ – A) = \sin A $

$ \sin 140^\circ = \sin ( 180^\circ – 40^\circ ) = \sin xl^\circ $

*). Menyelesaikan soal :

$ \begin{marshal} \sin 20^\circ \sin 40^\circ \sin 80^\circ & = \sin twoscore^\circ . (\sin lxxx^\circ . \sin 20^\circ ) \\ & = \sin 40^\circ . (\sin 80^\circ . \sin 20^\circ ) \\ & = \sin 40^\circ . \left( -\frac{i}{2} [ \cos (80^\circ + 20^\circ ) – \cos (lxxx^\circ – xx^\circ )] \right) \\ & = \sin 40^\circ . \left( -\frac{1}{2} [ \cos 100^\circ – \cos threescore^\circ ] \correct) \\ & = \sin xl^\circ . \left( -\frac{1}{ii} [ \cos 100^\circ – \frac{one}{ii} ] \correct) \\ & = \sin 40^\circ . \left( -\frac{one}{2} \cos 100^\circ + \frac{1}{4} \right) \\ & = -\frac{1}{ii} \cos 100^\circ \sin 40^\circ + \frac{i}{4} \sin 40^\circ \\ & = -\frac{one}{2} (\cos 100^\circ \sin 40^\circ ) + \frac{i}{4} \sin 40^\circ \\ & = -\frac{ane}{2} \times ( \frac{1}{2} [ \sin (100^\circ + xl^\circ ) – \sin (100^\circ – 40^\circ )] ) + \frac{1}{iv} \sin xl^\circ \\ & = ( -\frac{1}{iv} [ \sin 140^\circ – \sin sixty^\circ ] ) + \frac{1}{4} \sin 40^\circ \\ & = ( -\frac{i}{4} [ \sin 140^\circ – \frac{ane}{2}\sqrt{iii} ] ) + \frac{one}{4} \sin 40^\circ \\ & = -\frac{1}{iv} \sin 140^\circ + \frac{i}{8}\sqrt{3} + \frac{ane}{4} \sin 40^\circ \\ & = -\frac{1}{4} \sin forty^\circ + \frac{1}{8}\sqrt{3} + \frac{1}{iv} \sin 40^\circ \\ & = \frac{i}{8} \sqrt{3} \end{align} $

jadi, nilai $ \sin twenty^\circ \sin 40^\circ \sin lxxx^\circ = \frac{i}{8} \sqrt{3} . \, \heartsuit $.

2). Tentukan nilai dari bentuk

$ \cos 20^\circ \cos 40^\circ \cos 80^\circ $

?

Penyelesaian :

Ada empat cara yang akan kita sajikan dalam menyelesaikan soal nomor ii :

Cara I :

*). Rumus Dasar yang kita gunakan adalah rumus perkalian fungsi trigonometri :

$ \cos A . \cos B = \frac{1}{2} [ \cos (A+B) + \cos (A-B)] $

$ \cos ( 180^\circ – A) = -\cos A $

$ \cos 100^\circ = \cos ( 180^\circ – fourscore^\circ ) = -\cos 80^\circ $

*). Menyelesaikan soal :

$ \begin{align} \cos xx^\circ \cos 40^\circ \cos 80^\circ & = (\cos twenty^\circ \cos 40^\circ ) \cos eighty^\circ \\ & = (\cos twoscore^\circ \cos twenty^\circ ) \cos 80^\circ \\ & = \left( \frac{one}{ii} [ \cos (xl^\circ + 20^\circ ) + \cos (40^\circ – xx^\circ )] \correct) \cos lxxx^\circ \\ & = \left( \frac{ane}{ii} [ \cos 60^\circ + \cos 20^\circ ] \right) \cos lxxx^\circ \\ & = \left( \frac{1}{2} [ \frac{1}{2} + \cos 20^\circ ] \correct) \cos 80^\circ \\ & = \left( \frac{one}{4} + \frac{ane}{two} \cos twenty^\circ \right) \cos 80^\circ \\ & = \frac{1}{four} \cos 80^\circ + \frac{1}{ii} \cos 80^\circ \cos 20^\circ \\ & = \frac{1}{4} \cos 80^\circ + \frac{1}{ii} ( \cos 80^\circ \cos 20^\circ ) \\ & = \frac{1}{4} \cos 80^\circ + \frac{1}{2} \times ( \frac{1}{2} [ \cos (eighty^\circ + twenty^\circ ) + \cos (80^\circ – 20^\circ )] ) \\ & = \frac{i}{four} \cos 80^\circ + ( \frac{1}{4} [ \cos 100^\circ + \cos lx^\circ ] ) \\ & = \frac{1}{4} \cos 80^\circ + ( \frac{1}{four} [ -\cos 80^\circ + \frac{1}{ii} ] ) \\ & = \frac{1}{4} \cos 80^\circ -\frac{1}{4} \cos 80^\circ + \frac{1}{8} \\ & = \frac{ane}{8} \end{align} $

jadi, nilai $ \cos twenty^\circ \cos xl^\circ \cos 80^\circ = \frac{ane}{viii} . \, \heartsuit $.

Cara II :

*). Rumus Dasar yang kita gunakan adalah rumus perkalian fungsi trigonometri :

$ \cos A . \cos B = \frac{ane}{2} [ \cos (A+B) + \cos (A-B)] $

$ \cos ( 180^\circ – A) = -\cos A $

$ \cos 140^\circ = \cos ( 180^\circ – 40^\circ ) = -\cos 40^\circ $

*). Menyelesaikan soal :

$ \begin{align} \cos 20^\circ \cos 40^\circ \cos 80^\circ & = (\cos 80^\circ \cos 20^\circ ) \cos 40^\circ \\ & = (\cos 80^\circ \cos xx^\circ ) \cos 40^\circ \\ & = \left( \frac{one}{2} [ \cos (80^\circ + 20^\circ ) + \cos (fourscore^\circ – 20^\circ )] \correct) \cos forty^\circ \\ & = \left( \frac{one}{2} [ \cos 100^\circ + \cos 60^\circ ] \right) \cos 40^\circ \\ & = \left( \frac{1}{2} [ \cos 100^\circ + \frac{1}{2} ] \correct) \cos twoscore^\circ \\ & = \left( \frac{1}{2} \cos 100^\circ + \frac{ane}{4} \right) \cos forty^\circ \\ & = \frac{1}{4} \cos twoscore^\circ + \frac{1}{2} \cos 100^\circ \cos twoscore^\circ \\ & = \frac{ane}{4} \cos 40^\circ + \frac{1}{2} ( \cos 100^\circ \cos 40^\circ ) \\ & = \frac{1}{4} \cos 40^\circ + \frac{1}{ii} \times ( \frac{1}{2} [ \cos (100^\circ + forty^\circ ) + \cos (100^\circ – twoscore^\circ )] ) \\ & = \frac{ane}{iv} \cos 40^\circ + ( \frac{1}{4} [ \cos 140^\circ + \cos sixty^\circ ] ) \\ & = \frac{1}{4} \cos twoscore^\circ + ( \frac{1}{iv} [ -\cos forty^\circ + \frac{1}{2} ] ) \\ & = \frac{1}{iv} \cos 40^\circ -\frac{1}{4} \cos 40^\circ + \frac{ane}{viii} \\ & = \frac{i}{viii} \stop{align} $

jadi, nilai $ \cos twenty^\circ \cos xl^\circ \cos 80^\circ = \frac{one}{8} . \, \heartsuit $.

Baca :   Berikut Ini Yang Termasuk Unsur Logam Adalah

Cara III :

*). Rumus Dasar yang kita gunakan adalah rumus perkalian fungsi trigonometri :

$ \cos A . \cos B = \frac{1}{2} [ \cos (A+B) + \cos (A-B)] $

*). Menyelesaikan soal :

$ \brainstorm{align} \cos 20^\circ \cos 40^\circ \cos 80^\circ & = (\cos 80^\circ \cos xl^\circ ) \cos 20^\circ \\ & = \left( \frac{1}{2} [ \cos (lxxx^\circ + twoscore^\circ ) + \cos (eighty^\circ – forty^\circ )] \right) \cos twenty^\circ \\ & = \left( \frac{ane}{2} [ \cos 120^\circ + \cos 40^\circ ] \right) \cos 20^\circ \\ & = \left( \frac{i}{2} [ -\frac{one}{ii} + \cos twoscore^\circ ] \right) \cos twenty^\circ \\ & = \left( -\frac{1}{4} + \frac{one}{2} \cos 40^\circ \correct) \cos xx^\circ \\ & = – \frac{one}{4} \cos xx^\circ + \frac{i}{two} \cos 40^\circ \cos xx^\circ \\ & = – \frac{1}{4} \cos 20^\circ + \frac{i}{2} ( \cos 40^\circ \cos 20^\circ ) \\ & = – \frac{1}{4} \cos 20^\circ + \frac{1}{two} \times ( \frac{1}{2} [ \cos (40^\circ + 20^\circ ) + \cos (40^\circ – 20^\circ )] ) \\ & = – \frac{1}{4} \cos twenty^\circ + ( \frac{1}{4} [ \cos 60^\circ + \cos xx^\circ ] ) \\ & = – \frac{one}{4} \cos 20^\circ + ( \frac{i}{four} [ \frac{ane}{2} + \cos 20^\circ ] ) \\ & = – \frac{ane}{4} \cos twenty^\circ + \frac{1}{8} + \frac{1}{4} \cos twenty^\circ \\ & = \frac{i}{8} \terminate{marshal} $

jadi, nilai $ \cos 20^\circ \cos 40^\circ \cos fourscore^\circ = \frac{1}{eight} . \, \heartsuit $.

Cara 4 :

*). Rumus Dasar yang kita gunakan adalah “Rumus Trigonometri untuk Sudut Ganda” :

$ \sin 2A = ii\sin A \cos A \rightarrow \sin A \cos A = \frac{1}{two} \sin 2A $

Rumus Lain :

$ \sin (180^\circ – A) = \sin A $

$ \sin 160^\circ = \sin (180^\circ – 20^\circ ) = \sin 20^\circ $

*). Menyelesaikan soal ,

Kita misalkan hasilnya $ P $ atau $ \cos 20^\circ \cos xl^\circ \cos fourscore^\circ = P $ :

$ \begin{align} P & = \cos twenty^\circ \cos 40^\circ \cos 80^\circ \, \, \, \, \text{(kali } \sin 20^\circ ) \\ P . \sin 20^\circ & = \sin 20^\circ \cos twenty^\circ \cos xl^\circ \cos 80^\circ \\ & = (\sin xx^\circ \cos 20^\circ ) \cos twoscore^\circ \cos 80^\circ \\ & = ( \frac{one}{ii}\sin 40^\circ ) \cos xl^\circ \cos fourscore^\circ \\ & = \frac{i}{2}\sin 40^\circ \cos 40^\circ \cos 80^\circ \\ & = \frac{i}{2} ( \sin 40^\circ \cos 40^\circ ) \cos 80^\circ \\ & = \frac{1}{2} \times ( \frac{1}{2} \sin lxxx^\circ ) \cos 80^\circ \\ & = \frac{1}{iv} \sin 80^\circ \cos eighty^\circ \\ & = \frac{i}{iv} ( \sin lxxx^\circ \cos 80^\circ ) \\ & = \frac{1}{iv} \times ( \frac{1}{2} \sin 160^\circ ) \\ P . \sin xx^\circ & = \frac{1}{8} \sin twenty^\circ \, \, \, \, \text{(bagi } \sin twenty^\circ ) \\ P & = \frac{ane}{8} \finish{align} $

jadi, nilai $ \cos xx^\circ \cos xl^\circ \cos eighty^\circ = P = \frac{1}{8} . \, \heartsuit $.

3). Tentukan nilai dari
$ \cos twoscore^\circ + \cos eighty^\circ + \cos 160^\circ $
?

(Soal Un Matematika IPA tahun 2007)

Penyelesaian :

Soal ini bisa diselesaikan dengan berbagai cara, diantaranya :

Cara I :

*). Rumus dasar yang digunakan :

$ \cos A + \cos B = 2\cos \frac{(A+B)}{2} \cos \frac{(A-B)}{2} $

$ \cos (180^\circ – A ) = – \cos A $

Nilai $ \cos 160^\circ = \cos (180^\circ – 20^\circ ) = – \cos 20^\circ $

*). Menyelesaikan soal :

$ \brainstorm{align} \cos forty^\circ + \cos eighty^\circ + \cos 160^\circ & = ( \cos eighty^\circ + \cos 40^\circ ) + \cos 160^\circ \\ & = ( two\cos \frac{(80^\circ + 40^\circ )}{ii} \cos \frac{(lxxx^\circ – 40^\circ )}{2} ) + \cos 160^\circ \\ & = ( 2\cos \frac{(120^\circ )}{2} \cos \frac{(40^\circ )}{2} ) + (- \cos twenty^\circ ) \\ & = ( 2\cos sixty^\circ \cos 20^\circ ) – \cos 20^\circ \\ & = 2 . \frac{1}{2} \cos 20^\circ – \cos 20^\circ \\ & = \cos twenty^\circ – \cos xx^\circ \\ & = 0 \terminate{marshal} $

jadi, nilai $ \cos 40^\circ + \cos 80^\circ + \cos 160^\circ = 0 . \, \heartsuit $.

Baca :   Senyawa Alkohol Yang Jika Dioksidasi Menghasilkan Alkanon Adalah

Cara 2 :

*). Rumus dasar yang digunakan :

$ \cos A + \cos B = 2\cos \frac{(A+B)}{2} \cos \frac{(A-B)}{ii} $

*). Menyelesaikan soal :

$ \begin{marshal} \cos 40^\circ + \cos fourscore^\circ + \cos 160^\circ & = ( \cos 160^\circ + \cos lxxx^\circ ) + \cos 40^\circ \\ & = ( ii\cos \frac{(160^\circ + 80^\circ )}{2} \cos \frac{(160^\circ – eighty^\circ )}{2} ) + \cos 40^\circ \\ & = ( two\cos \frac{(240^\circ )}{two} \cos \frac{(80^\circ )}{2} ) + \cos xl^\circ \\ & = ( 2\cos 120^\circ \cos 40^\circ ) + \cos 40^\circ \\ & = 2 . -\frac{one}{2} \cos 40^\circ + \cos 40^\circ \\ & = – \cos twoscore^\circ + \cos 40^\circ \\ & = 0 \end{align} $

jadi, nilai $ \cos forty^\circ + \cos 80^\circ + \cos 160^\circ = 0 . \, \heartsuit $.

Cara III :

*). Rumus dasar yang digunakan :

$ \cos A + \cos B = 2\cos \frac{(A+B)}{two} \cos \frac{(A-B)}{ii} $

*). Menyelesaikan soal :

$ \brainstorm{align} \cos xl^\circ + \cos 80^\circ + \cos 160^\circ & = ( \cos 160^\circ + \cos 40^\circ ) + \cos fourscore^\circ \\ & = ( 2\cos \frac{(160^\circ + 40^\circ )}{ii} \cos \frac{(160^\circ – twoscore^\circ )}{2} ) + \cos 80^\circ \\ & = ( 2\cos \frac{(200^\circ )}{2} \cos \frac{(120^\circ )}{2} ) + \cos 80^\circ \\ & = ( 2\cos 100^\circ \cos sixty^\circ ) + \cos 80^\circ \\ & = ii . \cos 100^\circ . \frac{1}{2} + \cos 80^\circ \\ & = \cos 100^\circ + \cos lxxx^\circ \\ & = 2\cos \frac{(100^\circ + 80^\circ )}{2} \cos \frac{(100^\circ – 80^\circ )}{2} \\ & = two\cos \frac{180^\circ }{2} \cos \frac{twenty^\circ }{2} \\ & = two\cos 90^\circ \cos ten^\circ \\ & = 2 \times 0 \times \cos 10^\circ \\ & = 0 \end{align} $

jadi, nilai $ \cos forty^\circ + \cos 80^\circ + \cos 160^\circ = 0 . \, \heartsuit $.

4). Tentukan nilai dari $ \csc 10^\circ – \sqrt{iii} \sec 10^\circ $?

(soal SIMAK UI tahun 2013 Matematika IPA kode 133)

Penyelesaian :

*). Rumus dasar yang digunakan :

i). Sudut Rangkap :

$ \sin 2A = 2\sin A \cos A \rightarrow \sin A \cos A = \frac{1}{ii} \sin 2A $.

2). Selilisih sudut : $ \sin (A – B ) = \sin A \cos B – \cos A \sin B $.

three). Rumus lain :

$ \csc A = \frac{1}{\sin A} \, $ dan $ \sec A = \frac{i}{\cos A } $.

*). Menyelesaikan soal :

$ \begin{align} \csc 10^\circ – \sqrt{iii} \sec 10^\circ & = \frac{1}{\sin ten^\circ } – \frac{\sqrt{3} }{\cos ten^\circ } \\ & = \frac{1}{\sin 10^\circ } – \frac{\sqrt{3} }{\cos 10^\circ } \\ & = \frac{\cos x^\circ }{\sin 10^\circ \cos 10^\circ } – \frac{\sqrt{3} \sin 10^\circ }{\sin 10^\circ \cos 10^\circ } \\ & = \frac{\cos 10^\circ – \sqrt{3} \sin 10^\circ }{\sin 10^\circ \cos 10^\circ } \, \, \, \, \, \, \text{(modifikasi)} \\ & = \frac{ 2 \times ( \frac{i}{ii} . \cos x^\circ – \frac{1}{2} \sqrt{3} . \sin 10^\circ ) }{\sin 10^\circ \cos 10^\circ } \\ & = \frac{ two \times ( \sin 30^\circ . \cos x^\circ – \cos thirty^\circ . \sin x^\circ ) }{\frac{1}{2} . \sin 2 \times 10^\circ } \\ & = \frac{ 2 \sin ( thirty^\circ – 10^\circ ) }{\frac{1}{two} . \sin twenty^\circ } \\ & = \frac{ 4 \sin ( xx^\circ ) }{ \sin 20^\circ } \\ & = four \cease{marshal} $

Jadi, nilai dari $ \csc 10^\circ – \sqrt{iii} \sec 10^\circ = 4 . \, \heartsuit $.

       Demikian pembahasan materi
Penerapan Rumus Trigonometri pada Soal-soal Bagian 1
dan contoh-contohnya. Silahkan juga baca materi lain yang berkaitan dengan trigonometri.

Cos 80 Derajat

Source: https://www.konsep-matematika.com/2017/02/penerapan-rumus-trigonometri-pada-soal-soal-bagian-1.html

Check Also

Contoh Soal Perkalian Vektor

Contoh Soal Perkalian Vektor. Web log Koma – Setelah mempelajari beberapa operasi hitung pada vektor …