Ksm 2022

<p><strong>Soal:</strong><br>Pada segitiga lancip ABCABCABC diketahui:</p><p>BC=5BC = 5BC=5</p><p>AC=4AC = 4AC=4</p><p>cos⁡(A−B)=3132\cos(A - B) = \tfrac{31}{32}cos(A−B)=3231​</p><p>Ditanya: sin⁡A+sin⁡B\sin A + \sin BsinA+sinB</p><p>Langkah 1: Gunakan identitas cos selisih sudut</p><p>cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\cos(A - B) = \cos A \cos B + \sin A \sin Bcos(A−B)=cosAcosB+sinAsinB</p><p>Diketahui nilainya:</p><p>cos⁡Acos⁡B+sin⁡Asin⁡B=3132.\cos A \cos B + \sin A \sin B = \frac{31}{32}.cosAcosB+sinAsinB=3231​.&nbsp;</p><p>Langkah 2: Hubungan sisi dengan sinus</p><p>Dengan hukum sinus:</p><p>sin⁡Aa=sin⁡Bb=sin⁡Cc.\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}.asinA​=bsinB​=csinC​.</p><p>Sisi berhadapan:</p><p>a=BC=5  ⟹  sin⁡A=5ka = BC = 5 \implies \sin A = 5ka=BC=5⟹sinA=5k</p><p>b=AC=4  ⟹  sin⁡B=4kb = AC = 4 \implies \sin B = 4kb=AC=4⟹sinB=4k<br>untuk suatu konstanta k&gt;0k &gt; 0k&gt;0.</p><p>Jadi:</p><p>sin⁡A+sin⁡B=9k.\sin A + \sin B = 9k.sinA+sinB=9k.&nbsp;</p><p>Langkah 3: Gunakan identitas Pythagoras</p><p>cos⁡A=1−sin⁡2A=1−(5k)2,cos⁡B=1−(4k)2.\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - (5k)^2}, \quad \cos B = \sqrt{1 - (4k)^2}.cosA=1−sin2A​=1−(5k)2​,cosB=1−(4k)2​.</p><p>Karena segitiga lancip → semua cos positif.</p><p>Langkah 4: Masukkan ke persamaan cos(A - B)</p><p>cos⁡Acos⁡B+sin⁡Asin⁡B=3132.\cos A \cos B + \sin A \sin B = \frac{31}{32}.cosAcosB+sinAsinB=3231​.</p><p>Substitusi:</p><p>1−25k2⋅1−16k2+(5k)(4k)=3132.\sqrt{1 - 25k^2} \cdot \sqrt{1 - 16k^2} + (5k)(4k) = \frac{31}{32}.1−25k2​⋅1−16k2​+(5k)(4k)=3231​.&nbsp;</p><p>Langkah 5: Sederhanakan</p><p>(1−25k2)(1−16k2)+20k2=3132.\sqrt{(1 - 25k^2)(1 - 16k^2)} + 20k^2 = \frac{31}{32}.(1−25k2)(1−16k2)​+20k2=3231​.</p><p>Kuadratkan (hati-hati karena positif semua):</p><p>(1−25k2)(1−16k2)=(3132−20k2)2.(1 - 25k^2)(1 - 16k^2) = \left(\frac{31}{32} - 20k^2\right)^2.(1−25k2)(1−16k2)=(3231​−20k2)2.&nbsp;</p><p>Langkah 6: Hitung</p><p>Kiri:</p><p>1−41k2+400k4.1 - 41k^2 + 400k^4.1−41k2+400k4.</p><p>Kanan:</p><p>(3132−20k2)2=9611024−124032k2+400k4.\left(\frac{31}{32} - 20k^2\right)^2 = \frac{961}{1024} - \frac{1240}{32}k^2 + 400k^4.(3231​−20k2)2=1024961​−321240​k2+400k4. =9611024−38.75k2+400k4.= \frac{961}{1024} - 38.75k^2 + 400k^4.=1024961​−38.75k2+400k4.&nbsp;</p><p>Langkah 7: Samakan</p><p>1−41k2+400k4=9611024−38.75k2+400k4.1 - 41k^2 + 400k^4 = \frac{961}{1024} - 38.75k^2 + 400k^4.1−41k2+400k4=1024961​−38.75k2+400k4.</p><p>400k4400k^4400k4 hilang.</p><p>1−41k2=9611024−38.75k2.1 - 41k^2 = \frac{961}{1024} - 38.75k^2.1−41k2=1024961​−38.75k2. 1−9611024=41k2−38.75k2.1 - \frac{961}{1024} = 41k^2 - 38.75k^2.1−1024961​=41k2−38.75k2. 631024=2.25k2.\frac{63}{1024} = 2.25k^2.102463​=2.25k2. k2=631024×2.25=632304=7256.k^2 = \frac{63}{1024 \times 2.25} = \frac{63}{2304} = \frac{7}{256}.k2=1024×2.2563​=230463​=2567​. k=716.k = \frac{\sqrt{7}}{16}.k=167​​.&nbsp;</p><p>Langkah 8: Cari sin⁡A+sin⁡B\sin A + \sin BsinA+sinB</p><p>sin⁡A+sin⁡B=9k=9⋅716=9716.\sin A + \sin B = 9k = 9 \cdot \frac{\sqrt{7}}{16} = \frac{9\sqrt{7}}{16}.sinA+sinB=9k=9⋅167​​=1697​​.&nbsp;</p><p>✅ <strong>Jawaban:</strong></p><p>sin⁡A+sin⁡B=9716\sin A + \sin B = \frac{9\sqrt{7}}{16}sinA+sinB=1697​​</p>