Trigonometry! #49

Geometry Level 3

In a Δ A B C \Delta ABC the tangent of half the difference of two angles is one-third the tangent of half the sum of the angles. The ratio of the sides opposite to the angles is k : 2 k:2 . Find and enter the value of k k .

This problem is part of the set Trigonometry .


The answer is 1.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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2 solutions

Omkar Kulkarni
Feb 14, 2015

tan ⁡ ( B − C 2 ) = 1 3 tan ⁡ ( B + C 2 ) \tan \left(\frac{B-C}{2}\right)=\frac{1}{3}\tan\left(\frac{B+C}{2}\right)

tan ⁡ ( B − C 2 ) tan ⁡ ( B + C 2 ) = 1 3 \frac{\tan \left(\frac{B-C}{2}\right)}{\tan\left(\frac{B+C}{2}\right)}=\frac{1}{3}

b − c b + c = 1 3 \frac{b-c}{b+c}=\frac{1}{3}

b c = 1 2 \boxed{\frac{b}{c}=\frac{1}{2}}

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Raven Herd
Feb 13, 2015

According to napier 's formula, tan(B-C)/2=(b-c) /(b+c) cot A/2. where A,B,C belong to angles of triangle. Like above we can prove for other angles as well. Given, tan (B-C)/2=1/3 tan(B+C)/2 (b-c)/(b+c)cot(A/2)=1/3 cot(A/2)

(b-c)/(b+c) = 1/3 b/c=k/2 Applying componendo and dividendo , (b+c)/(b-c)=(k+2)/(k-2) 3=(k+2)/(k-2) Now solve for k by applying componendo and dividendo again.

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