Explanation
Step 1: Geometric Analysis using Exterior Angles
Let OP be the vertical tower of height h, where P is the top and O is the foot. The points lie on a straight horizontal line in the order O−C−B−A because the angle increases as you get closer to the tower (\alpha < 2\alpha < 3\alpha).
In △PAB: The exterior angle is ∠PBO=2α.
∠PBO=∠PAB+∠APB⟹2α=α+∠APB⟹∠APB=α
Since ∠PAB=∠APB=α, △PAB is an isosceles triangle. Therefore:
AB=PB
In △PBC: The exterior angle is ∠PCO=3α.
∠PCO=∠PBC+∠BPC⟹3α=2α+∠BPC⟹∠BPC=α
Step 2: Apply the Sine Rule in △PBC
Using the Sine Rule for sides BC and PB with their respective opposite angles:
sin(∠BPC)BC=sin(∠PCB)PB
Substitute the angles ∠BPC=α and ∠PCB=180∘−3α:
sinαBC=sin(180∘−3α)PB
Since sin(180∘−θ)=sinθ, we get:
sinαBC=sin3αPB
Rearranging the terms to isolate BCPB:
BCPB=sinαsin3α
Since we proved AB=PB in Step 1, we can replace PB with AB:
BCAB=sinαsin3α
Step 3: Trigonometric Simplification to Match the Option
Now, we apply the triple-angle identity for sine: sin3α=3sinα−4sin3α.
BCAB=sinα3sinα−4sin3α
Factor out sinα from the numerator:
BCAB=sinαsinα(3−4sin2α)
BCAB=3−4sin2α
Next, we use the double-angle cosine identity to convert sin2α into cos2α:
cos2α=1−2sin2α⟹2sin2α=1−cos2α⟹4sin2α=2(1−cos2α)
Substitute 4sin2α=2−2cos2α back into our equation:
BCAB=3−(2−2cos2α)
BCAB=3−2+2cos2α
BCAB=1+2cos2α
Correct Option Match:
1+2cos2α
