[tex]\displaystyle \text{Sin 45}^\circ = \frac{y}{\sqrt{20}}\\\\\frac{\sqrt{2}}{2} = \frac{y}{\sqrt{20}}\\\\\sqrt{2}\times\sqrt{20} = y \times 2\\\\\sqrt{40} = 2y\\\\2\sqrt{10} = 2y\\\\\sqrt{10} = y\\\\\textbf{Jadi, nilai dari y adalah } \sqrt{10} \textbf{ cm}[/tex]
Mencari nilai x
[tex]\displaystyle \text{Cos 45}^\circ = \frac{y}{\sqrt{20}}\\\\\frac{\sqrt{2}}{2} = \frac{y}{\sqrt{20}}\\\\\sqrt{2}\times\sqrt{20} = y \times 2\\\\\sqrt{40} = 2y\\\\2\sqrt{10} = 2y\\\\\sqrt{10} = y\\\\\textbf{Jadi, nilai dari x adalah } \sqrt{10} \textbf{ cm}[/tex]
Pembahasan :
Diketahui :
Ditanyakan : x dan y ???
Jawab :
[tex]\displaystyle \text{Sin 45}^\circ = \frac{y}{\sqrt{20}}\\\\\frac{\sqrt{2}}{2} = \frac{y}{\sqrt{20}}\\\\\sqrt{2}\times\sqrt{20} = y \times 2\\\\\sqrt{40} = 2y\\\\2\sqrt{10} = 2y\\\\\sqrt{10} = y\\\\\textbf{Jadi, nilai dari y adalah } \sqrt{10} \textbf{ cm}[/tex]
[tex]\displaystyle \text{Cos 45}^\circ = \frac{y}{\sqrt{20}}\\\\\frac{\sqrt{2}}{2} = \frac{y}{\sqrt{20}}\\\\\sqrt{2}\times\sqrt{20} = y \times 2\\\\\sqrt{40} = 2y\\\\2\sqrt{10} = 2y\\\\\sqrt{10} = y\\\\\textbf{Jadi, nilai dari x adalah } \sqrt{10} \textbf{ cm}[/tex]