Dua buah cermin disusun sehingga membentuk sudut dan menghasilkan n buah bayangan. Jika sudut kedua cermin dikurangi 20 ternyata banyaknya bayangan akan bertambah 3 buah. Tentukan sudut kedua cermin mula-mula
izharulhaq78
![\frac{360}{\theta}+m=n \\ \\ \frac{360}{\theta-20}+m=n+3 \\ \\ \theta=? \\ \\ \frac{360}{\theta-20}+m=n+3 \\ \frac{360}{\theta-20}+m= \frac{360}{\theta}+m+3 \\ \frac{360}{\theta-20}= \frac{360}{\theta}+3 \\ \frac{360}{\theta-20}-\frac{360}{\theta}=3 \\ \frac{360\theta-[360(\theta-20)]}{(\theta-20)(\theta)}=3 \\ \frac{360\theta-[360\theta-7200]}{\theta^2-20\theta}=3 \\360\theta-360\theta+7200=3(\theta^2-20\theta) \\7200=3\theta^2-60\theta \\ 0=3\theta^2-60\theta-7200](https://tex.z-dn.net/?f=+%5Cfrac%7B360%7D%7B%5Ctheta%7D%2Bm%3Dn+%5C%5C++%5C%5C+++%5Cfrac%7B360%7D%7B%5Ctheta-20%7D%2Bm%3Dn%2B3+%5C%5C++%5C%5C+%5Ctheta%3D%3F+%5C%5C++%5C%5C+%5Cfrac%7B360%7D%7B%5Ctheta-20%7D%2Bm%3Dn%2B3+%5C%5C+%5Cfrac%7B360%7D%7B%5Ctheta-20%7D%2Bm%3D+%5Cfrac%7B360%7D%7B%5Ctheta%7D%2Bm%2B3+%5C%5C+%5Cfrac%7B360%7D%7B%5Ctheta-20%7D%3D+%5Cfrac%7B360%7D%7B%5Ctheta%7D%2B3+%5C%5C+%5Cfrac%7B360%7D%7B%5Ctheta-20%7D-%5Cfrac%7B360%7D%7B%5Ctheta%7D%3D3+%5C%5C+%5Cfrac%7B360%5Ctheta-%5B360%28%5Ctheta-20%29%5D%7D%7B%28%5Ctheta-20%29%28%5Ctheta%29%7D%3D3+%5C%5C+%5Cfrac%7B360%5Ctheta-%5B360%5Ctheta-7200%5D%7D%7B%5Ctheta%5E2-20%5Ctheta%7D%3D3+%5C%5C360%5Ctheta-360%5Ctheta%2B7200%3D3%28%5Ctheta%5E2-20%5Ctheta%29+%5C%5C7200%3D3%5Ctheta%5E2-60%5Ctheta+%5C%5C+0%3D3%5Ctheta%5E2-60%5Ctheta-7200 "\frac{360}{\theta}+m=n \\ \\ \frac{360}{\theta-20}+m=n+3 \\ \\ \theta=? \\ \\ \frac{360}{\theta-20}+m=n+3 \\ \frac{360}{\theta-20}+m= \frac{360}{\theta}+m+3 \\ \frac{360}{\theta-20}= \frac{360}{\theta}+3 \\ \frac{360}{\theta-20}-\frac{360}{\theta}=3 \\ \frac{360\theta-[360(\theta-20)]}{(\theta-20)(\theta)}=3 \\ \frac{360\theta-[360\theta-7200]}{\theta^2-20\theta}=3 \\360\theta-360\theta+7200=3(\theta^2-20\theta) \\7200=3\theta^2-60\theta \\ 0=3\theta^2-60\theta-7200")
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