$a^{2}=b^{2}+c^{2}-2bc\cdot cos\, A$

$a^{2}=2^{2}+3^{2}-2\cdot2\cdot3\cdot\frac{1}{2}$$=4+9-6$$=7$

$a=\sqrt{7}$

$cos\, C=\frac{3^{2}+5^{2}-7^{2}}{2\cdot3\cdot5}$$=\frac{9+25-49}{30}$$=-\frac{15}{30}$$=-\frac{1}{2}$

sudut $c=120^{\circ}$

$7^{2}=8^{2}+9^{2}-2\cdot8\cdot9\cdot cos\, A$

$49=64+81-144\cdot cos\, A$

$144\, cos\, A=96$

$cos\, A=\frac{96}{144}=\frac{2}{3}$

$sin\, A=\frac{\sqrt{5}}{3}$

Gunakan aturan kosinus :

$x^{2}=4^{2}+6^{2}-2\cdot4\cdot6\cdot cos\,60^{\circ}$

$x^{2}=16+36-48\cdot\left(\frac{1}{2}\right)=28$

$x=\sqrt{28}=2\sqrt{7}$

$PR^{2}=PQ^{2}+QR^{2}$$-2\cdot PQ\cdot QR\cdot cos\, Q$

$PR^{2}=6^{2}+8^{2}-2\cdot6\cdot8\cdot cos\,120^{\circ}$

$PR^{2}=36+64-2\cdot6\cdot8\cdot\left(-\frac{1}{2}\right)$$=100+48=148$

$PR=\sqrt{148}$$=2\sqrt{37}$

$c^{2}=a^{2}+b^{2}-2\cdot a\cdot b\cdot cos\, C$

$7^{2}=3^{2}+5^{2}-2\cdot3\cdot5\cdot cos\, C$

$49=9+25-30\cdot cosC$

$30\cdot cos\, C=-15$

$cos\, C=-\frac{15}{30}$$=-\frac{1}{2}$

$cos\, C=cos\,120^{\circ}$

$C=120^{\circ}$

Jadi $tan\, C=tan\,120^{\circ}$$=tan\,(180-60)^{\circ}$$=-tan\,60^{\circ}$$=-\sqrt{3}$

Perhatikan gambar berikut!

$AB^{2}=AC^{2}+BC^{2}-2\cdot AC\cdot BC\cdot cos\, C$

$AB^{2}=\left(2a\right)^{2}+\left(2a\sqrt{3}\right)^{2}-2\cdot\left(2a\right)\cdot\left(2a\sqrt{3}\right)\cdot cos\,30^{\circ}$

$AB^{2}=4a^{2}+12a^{2}-8a^{2}\sqrt{3}\cdot\left(\frac{1}{2}\sqrt{3}\right)$$=16a^{2}-12a^{2}=4a^{2}$

$AB=\sqrt{4a^{2}}=2a$

Gunakan aturan kosinus untuk mencari panjang sisi $PR$

$PR^{2}=PQ^{2}+QR^{2}-2\cdot PQ\cdot QR\cdot cos\, Q$

$PR^{2}=6^{2}+9^{2}-2\cdot6\cdot9\cdot cos\,120^{\circ}$

$PR^{2}=36+81-108\left(-\frac{1}{2}\right)$$=117+54$$=171$

$PR=\sqrt{171}$$=3\sqrt{19}$

Keliling segitiga PQR :

$PQ+QR+PR=6+9+3\sqrt{19}$$=\left(15+3\sqrt{19}\right)$ cm

Perhatikan gambar berikut!

$L\,\triangle ABC=\frac{1}{2}\cdot AB\cdot BC\cdot sin\, A$

$128=\frac{1}{2}\cdot16\cdot16\cdot sin\, A$

$128=128\cdot sin\, A$

$sin\, A=1$

$A=90^{\circ}$

Berarti segitiga ABC siku-siku di A. Gunakan phytagoras untuk mencari sisi BC

$BC=\sqrt{AB^{2}+AC^{2}}$$=\sqrt{16^{2}+16^{2}}$$=\sqrt{256+256}$$=\sqrt{512}$$=16\sqrt{2}$

Perhatikan gambar berikut ini!

$cos\, x^{\circ}=\frac{AB^{2}+AC^{2}-BC^{2}}{2.AB\cdot BC}$

$cos\, x^{\circ}=\frac{8^{2}+15^{2}-13^{2}}{2\cdot8\cdot15}$$=\frac{64+225-169}{240}$$=\frac{120}{240}=\frac{1}{2}$

$x^{\circ}=60^{\circ}$

Jadi $sin\, x^{\circ}\cdot tan\, x^{\circ}=sin\,60^{\circ}\cdot tan\,60^{\circ}=\frac{1}{2}\sqrt{3}\cdot\sqrt{3}=\frac{3}{2}$

Perhatikan gambar berikut!

$cos\, B=\frac{a^{2}+c^{2}-b^{2}}{2\cdot a\cdot c}$

$cos\, B=\frac{5^{2}+6^{2}-4^{2}}{2\cdot5\cdot6}$$=\frac{25+36-16}{60}$$=\frac{45}{60}$$=\frac{3}{4}$

$tan\, B=\frac{1}{3}\sqrt{7}$

$b+c=13$…..(1)

$a+c=12$$\rightarrow c=12-a$….(2)

Persamaan (1) dikurangkan dengan persamaan (2) sehingga diperoleh :

$b-a=1$$\rightarrow b=a+1$….(3)

Gunakan aturan kosinus :

$a^{2}=b^{2}+c^{2}-2bc\cdot cos\,60^{\circ}$

$a^{2}=\left(a+1\right)^{2}+\left(12-a\right)^{2}$$-2\left(a+1\right)\left(12-a\right)\cdot\frac{1}{2}$

$a^{2}=a^{2}+2a+1+144$$-24a+a^{2}-\left(11a-a^{2}+12\right)$

$a^{2}=a^{2}+2a+1+144$$-24a+a^{2}-11a+a^{2}-12$

$2a^{2}-33a+133=0$

$\left(2a-19\right)\left(a-7\right)=0$

$a=\frac{19}{2}$ atau $a=7$

Jika $a=7$, maka $b=8$ dan $c=5$

Jadi nilai $a^{2}+b^{2}+c^{2}=7^{2}+8^{2}+5^{2}$$=49+64+25$$=138.$

$\left(a-b\right)\left(a+b\right)=c\left(c-b\sqrt{3}\right)$

$a^{2}-b^{2}=c^{2}-bc\sqrt{3}$

$a^{2}=b^{2}+c^{2}-bc\sqrt{3}$. . . (1)

Ingat aturan kosinus :

$a^{2}=b^{2}+c^{2}-2bc\cdot cos\, A$. . . (2)

Substitusikan pers (1) ke pers (2) :

$b^{2}+c^{2}-bc\sqrt{3}=b^{2}+c^{2}-2bc\cdot cos\, A$

$2bc\cdot cos\, A=bc\sqrt{3}$

$cos\, A=\frac{1}{2}\sqrt{3}$

$cos\, A=co\, s30^{\circ}$

$A=30^{\circ}$

$\left(a+b+c\right)$$\left(b+c-a\right)=3bc$

$\left[\left(b+c\right)+a\right]$$\left[\left(b+c\right)-a\right]=3bc$

$\left(b+c\right)^{2}-a^{2}=3bc$

$b^{2}+c^{2}+2bc-a^{2}=3bc$

$(b^{2}+c^{2}-a^{2})=bc$

$2bc\cdot cos\,\alpha=bc$

$bc\left(2\cdot cos\,\alpha-1\right)=0$

$bc=0$atau $cos\,\alpha=\frac{1}{2}$

Jadi $\alpha=60^{\circ}$

Gunakan aturan kosinus pada segitiga $ABD:$

$cos\, A=\frac{c^{2}+\frac{b^{2}}{4}-d^{2}}{2(c)\left(\frac{b}{2}\right)}$…..(1)

Aturan kosinus pada segitiga $ABC:$

$cos\, A=\frac{c^{2}+b^{2}-a^{2}}{2(b)(c)}$….(2)

Substitusikan pers (1) ke pers (2) :

$\frac{c^{2}+\frac{b^{2}}{4}-d^{2}}{2(c)\left(\frac{b}{2}\right)}=\frac{c^{2}+b^{2}-a^{2}}{2(b)(c)}$

$2c^{2}+\frac{b^{2}}{2}-2d^{2}=c^{2}+b^{2}-a^{2}$

$2d^{2}=c^{2}-\frac{1}{2}b^{2}+a^{2}$

$d^{2}=\frac{1}{2}c^{2}-\frac{1}{4}b^{2}+\frac{1}{2}a^{2}$