MathLabs

Problem 3

In a given right triangle ABCABC, the hypotenuse BCBC has length aa and is divided into nn equal parts, where nn is odd. The central part subtends an angle α\alpha at AA. If hh is the perpendicular distance from AA to BCBC, prove that tan⁡α=4nha(n2−1)\tan\alpha=\dfrac{4nh}{a(n^2-1)}.
Step 5 of 5: Simplify to the required identity
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tan⁡α=ah/na2/4−a2/(4n2)=4nha(n2−1)\tan\alpha=\frac{a h/n}{a^2/4-a^2/(4n^2)}=\frac{4nh}{a(n^2-1)}
Illustrative angle: for an isosceles right example with a=2,h=1,n=3a=2,h=1,n=3, the formula gives tan⁡α=3/4\tan\alpha=3/4.
Unit-circle illustration of a 36.87-degree angle, an illustrative numerical value for the final tangent identity.
Detailed analysis

Substitute PQ=a/nPQ=a/n, AH=hAH=h, the product from Step 3, and h2+MH2=a2/4h^2+MH^2=a^2/4 into Step 2. The denominator becomes a2/4−a2/(4n2)=a2(n2−1)/(4n2)a^2/4-a^2/(4n^2)=a^2(n^2-1)/(4n^2), so tan⁡α=4nh/[a(n2−1)]\tan\alpha=4nh/[a(n^2-1)].