Trigonometry F.4 ( not A. Maths ) *30marks*

(a) a: b: c = sin40∘: sin60∘: sin80∘
so a = c sin40∘/sin80∘.............................for part (c)

(b) AB is the longest

(c) a c sin60∘/ 2 = 80
c sin60∘c sin40∘/sin80∘= 160
c^2 = 160 sin 80∘/ sin40∘sin60∘
c = sqrt (160 sin 80∘/ sin40∘sin60∘)
c = 16.824cm
So AB = 16.8cm (corr. to 3s.f.)

(a) Find the ration a:b:c in terms of sines of angles
a/sin40 = b/sin60 = c/sin80
so, a:b:c = sin40 : sin60 : sin80
(b) Which side has the longest length
c
(c) Hence find the length of the
area = 80 = 0.5 (ksin40)(ksin60)sin80 = 0.5*k^2*sin40sin60sin80
160/sin40sin60sin80 = k^2
k = 17.1 cm
So, longest side = 17.1 * sin 80 = 16.8 cm

(a) a/sin40 = b/sin60 = c/sin80, where a=BC, b=AC, c=AB
(b) sin40< sin60 < sin80, therefore, a>b>c.
i.e. BC (a) is the longest.
(c) Area = 80 = 0.5 * a * b * sin80
a^2 = 160sin40/(sin80sin60) (b= asin60/sin40)
a = 10.98 cm