DIKETAHUI
[tex]\sf 2.~cos~\alpha=\dfrac{5}{13},~\alpha~sudut~lancip[/tex]
[tex]\sf 3.~sin~A=\dfrac{12}{13}[/tex]
[tex]\sf 4.~sin~60^o+tan~30^o+cos~45^o[/tex]
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DITANYA
[tex]\sf 2.~2~sin~\alpha=...[/tex]
[tex]\sf 3.~2~sin~A~cos~A=...[/tex]
[tex]\sf 4.~Nilainya=...[/tex]
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JAWAB
[tex]\begin{aligned}\sf 2.~2~sin~\alpha&=\sf 2 \times \dfrac{ \sqrt{ {13}^{2} - {5}^{2} }}{13} \\&=\sf 2 \times \dfrac{ \sqrt{169-25}}{13} \\&=\sf 2 \times \dfrac{ \sqrt{144}}{13} \\&=\sf 2 \times \dfrac{ 12}{13} \\&=\sf\dfrac{ 24}{13}\end{aligned}[/tex]
[tex]\begin{aligned}\sf 3.~2~sin~A~cos~A&=\sf 2 \times \dfrac{12}{13} \times\dfrac{ \sqrt{ {13}^{2} - {12}^{2} }}{13} \\&=\sf \dfrac{24}{13} \times \dfrac{ \sqrt{169-144}}{13} \\&=\sf \dfrac{24}{13} \times \dfrac{ \sqrt{25}}{13} \\&=\sf \dfrac{24}{13} \times \dfrac{5}{13} \\&=\sf\dfrac{120}{169}\end{aligned}[/tex]
[tex]\begin{aligned}\sf 4.~sin~60^o+tan~30^o+cos~45^o&=\sf \dfrac{1}{2} \sqrt{3} + \dfrac{1}{3} \sqrt{3} + \dfrac{1}{2} \sqrt{2}\\&=\sf \dfrac{\sqrt{3}}{2}+ \dfrac{\sqrt{3}}{3}+ \dfrac{\sqrt{2}}{2}\\&=\sf \dfrac{3\sqrt{3}}{6}+ \dfrac{2\sqrt{3}}{6}+ \dfrac{3\sqrt{2}}{6}\\&=\sf \dfrac{5\sqrt{3} +3 \sqrt{2} }{6}\end{aligned}[/tex]
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