An isosceles triangle is inscribed in a circle. The shortest side is the base which is 16 cm long. If the radius of the circle is 10 cm, what is the length of side "a"?
Put your answer to 1 decimal place into answerblank 1 (no spaces)
Answer:
draw a diagram. Let AB be the short side of the triangle, and let C be the center of the circle. Let D be the angle opposite AB.
AC = BC = 10 and AB = 16
If we let θ = angle ABC, then γ = angle ACB = 180-θ-θ = 180-2θ
AC/sinθ = AB/sin(180-2θ)
now, sin(180-2θ) = sin2θ, so
10/sinθ = 16/sin2θ
20sinθcosθ = 16sinθ, so
cosθ = 4/5 and sinθ = 3/5
so, sin2θ = 24/25
Now, if we let α = angle CAD = angle CBD, and β = angle ADB, we have
(θ+α) + (θ+α) + β = 180° in triangle ABD. Now, γ, being a central angle is 2β, so β = γ/2. Since 2θ+γ=180, γ/2=90-θ, and we have
2θ+2α + 90-θ = 180
θ+2α = 90
2α=90-θ
so, sin2α = cosθ
and sinα = 1/√10, cosα = 1/3
back to trangle ABD, we want side a, which is found using
a/sin(θ+α) = 16/sinβ
using the values above, we have
sinsin(θ+α) = (1+4√10)/5 and sinβ = 24/25, so
a/sin(θ+α) = 50/3
a = 50/3((1+4√10)/5) = 5(1+4√10)
whew - better check my math.