mohon bantuannya.. ini saya kurang paham

[tex]29) \:30 \div 60 \div 90 = 1 \div \sqrt{3} \div 2 = qr \div pr \div pq \\ karena \: pr = 2 \sqrt{2} \: maka \: pq = \frac{4}{3} \sqrt{6} [/tex]
[tex]30) \: {c}^{2} = {a}^{2} + {b}^{2} - 2ab \cos(c) \\ {c}^{2} = {5}^{2} + {4}^{2} - 2.5.4. \cos(60) \\ {c}^{2} = 25 + 16 - 40. \frac{1}{2} \\ {c}^{2} = 41 - 20 = 21 \\ c = \sqrt{21} [/tex]