Dua buah vektor gaya memiliki besar yang sama, yaitu 10 N. Perbandingan antara resultan dan selisih kedua vector adalah 3. Besar sudut apit kedua vector gaya ini adalah …
a. 30°
b. 37°
c. 35°
d. 60°
e. 90°
Guru Menjawab \begin{aligned}\frac{\left|\vec{A}+\vec{B}\right|}{\left|\vec{A}-\vec{B}\right|}&=&\frac{\sqrt{\vec{A^2}+\vec{B^2}+2.\vec{A.}\vec{B}\cos{\alpha}}}{\sqrt{\vec{A^2}+\vec{B^2}-2.\vec{A.}\vec{B}\cos{\alpha}}}\\ 3&=&\frac{\sqrt{\vec{{10}^2}+\vec{{10}^2}+2.\vec{10.}\vec{10}\cos{\alpha}}}{\sqrt{\vec{{10}^2}+\vec{{10}^2}-2.\vec{10.}\vec{10}\cos{\alpha}}}\\3&=&\frac{\sqrt{200+200\cos{\alpha}}}{\sqrt{200-200\cos{\alpha}}}\\9&=&\frac{200+200\cos{\alpha}}{200-200\cos{\alpha}}\\18-18\ cos\alpha&=&2+2\cos{\alpha}\\ 20\ cos\alpha&=&16\\ \ cos\alpha&=&0,8\\ cos&=&37°\end{aligned}
a. 30°
b. 37°
c. 35°
d. 60°
e. 90°
Guru Menjawab \begin{aligned}\frac{\left|\vec{A}+\vec{B}\right|}{\left|\vec{A}-\vec{B}\right|}&=&\frac{\sqrt{\vec{A^2}+\vec{B^2}+2.\vec{A.}\vec{B}\cos{\alpha}}}{\sqrt{\vec{A^2}+\vec{B^2}-2.\vec{A.}\vec{B}\cos{\alpha}}}\\ 3&=&\frac{\sqrt{\vec{{10}^2}+\vec{{10}^2}+2.\vec{10.}\vec{10}\cos{\alpha}}}{\sqrt{\vec{{10}^2}+\vec{{10}^2}-2.\vec{10.}\vec{10}\cos{\alpha}}}\\3&=&\frac{\sqrt{200+200\cos{\alpha}}}{\sqrt{200-200\cos{\alpha}}}\\9&=&\frac{200+200\cos{\alpha}}{200-200\cos{\alpha}}\\18-18\ cos\alpha&=&2+2\cos{\alpha}\\ 20\ cos\alpha&=&16\\ \ cos\alpha&=&0,8\\ cos&=&37°\end{aligned}
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