Ex10C 3.5.10.13

(13 a) ∠ABC = 100, ∠BAC = 20
So, by sine law:
AC/sin ∠ABC = BC/sin ∠BAC
AC = 60 (sin 100/sin 20) = 172.8 m
(13 b) RC/BC = tan 16.7
RC = 60 tan 16.7 = 18.0 m
(13 c) ∠ACB = 180 - 20 - 100 = 60
Area = 0.5 (AC)(BC) sin ∠ACB
= 0.5 x 172.8 x 60 sin 60
= 4489 m2
(10 a) Using pyth. thm:
VO2 = VA2 - OA2
= 64
VO = 8 cm
(10 b) The required angle is equal to that between VA and ABC, so,
cos ∠VAO = 0.6
∠VAO = 53.1
(6 a) TP/PQ = tan ∠TQP
TP = 65 tan 50 = 77.5 m
(6 b) TP/PR = tan ∠TRP
PR = 77.5/tan 25 = 166.1 m
So,
tan ∠QRP = QP/RP = 0.391
∠QRP = 21.4
So the reduced bearing of Q from R is N68.4E.
(3 a) AH2 = AD2 + DH2
= AD2 + DC2 + CH2
= 50
AH = 7.07 m
(3 b) sin ∠AHD = AD/AH = 0.424
∠AHD = 25.1

1.
60/sin20 * sin 100

2.
tan 16.7 *60

3.
sin60 *60 *(ans in a) *1/2