急!F5 Solving Trinages 10q2

7.
∠ABC + ∠BCA + ∠CAB = 180°
3∠BCA = 180° since ABC is an equilateral triangle
∠BCA = 60°
i.e.
∠ACD
= ∠DCB - ∠BCA
= 90° - 60°
= 30°

(10.1)/sin∠ACD = 20/sin∠DAC
(10.1)/(1/2) = 20/sin∠DAC
sin∠DAC = 20/20.2 = 100/101
since 0° < ∠DAC < 180°
∠DAC = 81.93° or (180° - 81.93°)
∠DAC = 81.93° or 98.07°

25.
Obviously,
C = 180° - 50° - 60° = 70°

By the Sine Rule:
Let BC/sinA = CA/sinB = AB/sinC = k
Area = (1/2)(AB)(BC)sinB = 82
(1/2)(ksinC)(ksinA)sinB = 82
(k^2)(sin70°)(sin50°)sin60° = 164
k^2 = 164/(sin70°)(sin50°)sin60° = 263.07
k = 16.22
°) - 3cos20°
i.e.
AB = ksinC = (16.22)(sin70°) = 15.24 cm
BC = ksinA = (16.22)(sin50°) = 12.42 cm
CA = ksinB = (16.22)(sin60°) = 14.05 cm

28.
(a)
PT
= [PB^2 + BT^2 - 2(PB)(BT)cos110°]^(1/2)
= (41 - 40cos110°)^(1/2)
= 7.39 cm

(b)
4/sin∠BPT = (7.39)/sin∠PBT
sin∠BPT = 4sin110°/7.39 = 0.51
∠PAB = ∠BPT (∠s in alt. segment)
sin∠PAB = 0.51
∠PAB = 30.55°

(c)
AP/sin∠ABP = PB/sin∠PAB
AP
= 5sin(180° - 110°)/0.51
= 5sin70°/0.51
= 9.24 cm