Prosiłabym o pomoc z zadaniem 20 i 18 ze zdjęcia

20. T: |AB| = |BC|
D: sin30st. = |AB|/6 <=> 1/2 = |AB|/6 <=> |AB| = 3
Z tw. Pitagorasa: 6^2 + |BC|^2 = (3√5)^2
|BC|^2 = 45 - 36 = 9, |BC| > 0, zatem |BC| = 3
|AB| = |BC|, cnw
18. a = 5cm, tg60st. = h/a <=> √3 = h/5 <=> h = 5√3 cm
6a + 3h = 6 * 5 + 3 * 5√3 = (30 + 15√3) cm