有關cos sin tan 既問題

Draw a perpendicular bisector(垂直平分線) AD such that BD = DC and AD is perpendicular to BC.
D is a point on the line BC.
Let AB = AC = x
AB + AC + BC = 10 m
x + x + BC = 10
BC = 10 – 2x
BD = ½ BC = ½ (10 – 2z) = 5 –x

Sine angle ABD = sin 30º = AD/AB
AD = AB sin 30º
AD = x sin 30º = x (0.5) = ½ x

Consider triangle ABD (right angle triangle)
AB^2 = BD^2 + AD^2 (Pythagorean theorem)畢達哥拉斯定理
x^2 = (5 –x)^2 + (½ x)^2
x^2 = 25 –10x + x^2 + 1/4 x^2
1/4 x^2 – 10x + 25 = 0
x^2 – 40x + 100 = 0
Using Quadratic formula,(二次方程式)
x = [-b+-(b^2 - 4ac)^0.5]/2
x = {-(-40)+-[(-40)^2 - 4(100)]^0.5}/2
x = {40+-[1200]^0.5}/2
x = {40+-34.641}/2
x =37.32 or 2.6795
Drop 37.32
x = 2.6795 m
BC = 10 – 2(2.6795 m) = 4.641 m

For triangle ABC
AB = 2.6795 m, AC = 2.6795 m, BC = 4.641 m (accurate up to 4 decimal place)
angle A = 120º, angle B = angle C = 30º

cos, tan and sin are used in a right-angled triangle