Answer:
d = k·sin(2θ)·sin(α)/(sin(θ)·sin(β))
Step-by-step explanation:
The Law of Sines tells us that sides of a triangle are proportional to the sine of the opposite angle. This can be used along with a trig identity to demonstrate the required relation.
__
<h3>top triangle</h3>
The law of sines applied to the top triangle is ...
BC/sin(A) = AC/sin(θ)
Triangle ABC is isosceles, so the base angles at B and C are congruent. Then the angle at vertex A is ...
∠A = 180° -θ -θ = 180° -2θ
A trig identity tells us the sine of an angle is equal to the sine of its supplement. That means the sine of angle A is ...
sin(A) = sin(180° -2θ) = sin(2θ)
and our above Law of Sines equation tells us ...
BC = sin(A)/sin(θ)·AC = k·sin(2θ)/sin(θ)
__
<h3>bottom triangle</h3>
The law of sines applied to the bottom triangle is ...
DC/sin(B) = BC/sin(D)
d/sin(α) = BC/sin(β)
Multiplying by sin(α) we have ...
d = BC·sin(α)/sin(β)
__
Using our expression for BC gives the desired relation:
d = k·sin(2θ)·sin(α)/(sin(θ)·sin(β))
