急! F5 Trigonometry 11q6

18a)
by cosine law,
AB²
=AN²+NB² - 2(AN)(NB)cos∠ANB
=... (代返d數計就得)

AB = 5sqrt(2-sqrt3)cm (用計數機take 3 sig fig)


18b)
AC²=AB²+BC²
...(代返d數計就得)
AC= sqrt(66-25sqrt3)cm (用計數機take 3 sig fig)

18c)
NM = 2 (mid-pt thm)
AM = sqrt29 (pyth thm)
AC = 2sqrt29 (corr. sides, ~Δs)
so MC = sqrt29
cosθ = (AM² + MC² - AC²) / [2(AM)(MC)] (cosine law)
...(代返d數計就得)
θ = 52.5'


19a)
ST² = PT² + SP² (pyth thm)
...
ST = 25cm

19b)
PT=QT and PS=QR, so TS=TR, ie ΔTSR is isosceles triangle
so ∠TMS = 90' (prop. of isos.Δ)
so TM² = ST² - MS² (pyth thm)
...
TM = 24cm

19c)
let N be the mid-pt of PQ
required angle = ∠MTN
sin∠MTN = MN/MT = 15/24
∠MTN = 38.7'

19d)
required angle is ∠NMT
∠NMT = 180' - ∠MNT - ∠MTN (∠ sum of Δ)
...
∠NMT = 51.3'

19e)
required angle is ∠QPT
cos∠QPT = NP/PT = 7/20 (since ΔPQT is isosceles triangle, PN=NQ)
∠QPT = 69.5'