triangle problem

Ques. 1:
Angle PRQ = 180 – 75 – 45 = 60 deg
Using sine rule,
PR/sin 45 = PQ/sin 60
PR/PQ = sin 45/sin 60 = (1/sq. rt 2)/[(sq. rt3) /2] = 2/sq.rt.6 = (sq.rt 6)/3
PR/PQ = (sq rt6) /3

Note: sq.rt 6 = square root of 6

Ques. 2
Draw a vertical line BE perpendicular to DC, such that E is on DC
Draw a vertical line AF perpendicular to DC, such that F is on DC
Consider triangle BCE
sin 72 deg =BE/BC
0.9511 =BE/3.6 cm
BE = 3.6 cm (0.9511) = 3.4238 cm
AF = BE = 3.4238 cm
Consider triangle ADF
sin 65 deg =AF/AD
0.9063 = 3.4238 cm /AD
AD = 3.4238 cm /0.9063
AD = 3.7778 cm
Tan 65 deg = AF/DF
2.1445 = 3.4238 cm /DF
DF = 3.4238 cm/2.1445 = 1.5965 cm
Tan 72 deg = BE/EC
3.07777 = 3.4238 cm/EC
EC = 3.4238 cm/3.07777 = 1.1125 cm
DC = DF + FE + EC = 1.5965 cm + 3.5 cm + 1.1125 cm = 6.209 cm
(a) AD = 3.7778 cm (b) DC = 6.209 cm

Ques. 3.
Consider triangle ABD
Using sine rule
Sin 30 deg/DB = sin angle DBA/DA
0.5/6 cm = sin angle DBA/9 cm
sin angle DBA = (0.5/6)(9)= 0.75
angle DBA = sin^-1(0.75) = 48.59 deg or 131.41 deg
Angle DBA = 131.41 deg and angle DBC = 48.59 deg
Consider isosceles triangle CDB,
CD = BD = 6 cm, angle DBC = angle BCD = 48.59 deg
Angle BDC = 180 -48.59 - 48.59 = 82.8 deg
Draw a vertical line DE to BC, DE is a perpendicular bisector,
Consider triangle DBE
Cos angle EDE = EB/DB
Cos 48.59 deg = EB/6 cm
EB = 6cm cos 48.59 deg = 6 cm (0.6614) = 3.9687 cm
BC = CE + EB = 3.9687 cm + 3.9687 cm = 7.373 cm
(a) angle BDC = 82.8 deg (b) BC = 7.373 cm