急! F5 Trigonometry 11q1

10.
∠VDM = 75° (Given)
i.e.
DM/VD = cos75°
DM
= 10cos(30° + 45°)
= 10[cos30°cos45° - sin30°sin45°)
= 10{[(3)^(1/2)/2][1/2^(1/2)] - (1/2)[1/2^(1/2)]}
= 10[3^(1/2) - 1]/2(2)^(1/2)
= 5[3^(1/2) - 1]/(2)^(1/2)
= 5(2)^(1/2)[3^(1/2) - 1]/2
= (5/2)[6^(1/2) - 2^(1/2)] cm

DB
= 2DM
= 5[6^(1/2) - 2^(1/2)] cm

AD^2 + AB^2 = DB^2
2AD^2 = 25[6 - 2(12)^(1/2) + 2]
AD^2 = (25/2)[8 - 4(3)^(1/2)] = 25[4 - 2(3)^(1/2)] = 25[3^(1/2) - 1]^2
AD = 5[3^(1/2) - 1] cm , since AD > 0
i.e.
Length of a side of the base = 5[3^(1/2) - 1] cm.

if surd form is not necessary:
∠VDM = 75° (Given)
i.e.
DM/VD = cos75°
DM = 10cos75° = 2.59 cm
DB = 2DM = 5.18 cm
AD^2 + AB^2 = (5.18)^2 = 26.79
AD^2 = 13.40
AD = 3.66 cm , since AD > 0
i.e.
Length of a side of the base = 3.66 cm.

14.
(a)
tan x = DA/AE = CB/BE = CB/(12 - AE)
(12 - AE) = AE(CB/DA) = AE(5/10) = AE/2
24 - 2AE = AE
AE = 8 cm

i.e. tan x = DA/AE = 10/8 = 5/4
x = 51.34°

(b)
DE
= (AD^2 + AE^2)^(1/2)
= (100 + 64)^(1/2)
= 2(41)^(1/2)

(c)
CE
= (CB^2 + BE^2)^(1/2)
= (5^2 + 4^2)^(1/2)
= (41)^(1/2) cm

CD
= (CA^2 + AD^2)^(1/2)
= [(CB^2 + BA^2) + AD^2]^(1/2)
= [(5^2 + 12^2) + 10^2]^(1/2)
= (269)^(1/2) cm

By Cosine Law,
cos∠CED
= (CE^2 + ED^2 - CD^2)/2(CE)(ED)
= (41 + 164 - 269)/2[(41)^(1/2)][2(41)^(1/2)]
= -64/4(41)
= -16/41
∠CED = 112.97°