Tangent ratio

Geometry Level 3

In △ A B C \triangle ABC , the corresponding side lengths opposite angles A A , B B , and C C are a a , b b , and c c respectively. If a cos ⁡ B − b cos ⁡ A = 3 5 c a\cos B-b\cos A=\dfrac{3}{5}c , find the value of tan ⁡ A tan ⁡ B \dfrac{\tan A}{\tan B} .


The answer is 4.

Danish Ahmed by Danish Ahmed , 24, India

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4 solutions

According to the Extended Sine Rule , a = 2 R sin ⁡ A b = 2 R sin ⁡ B c = 2 R sin ⁡ C a=2R \sin A \ \ \ \ \ b=2R \sin B \ \ \ \ \ c=2R \sin C where R R is the circumradius of △ A B C \triangle ABC .

Hence we have

a cos ⁡ B − b cos ⁡ A = 3 5 c ⇔ 2 R sin ⁡ A cos ⁡ B − 2 R sin ⁡ B cos ⁡ A = 3 5 2 R sin ⁡ C ⇔ sin ⁡ A cos ⁡ B − sin ⁡ B cos ⁡ A = 3 5 sin ⁡ C ⇔ sin ⁡ A cos ⁡ B − sin ⁡ B cos ⁡ A = 3 5 sin ⁡ ( A + B ) ⇔ sin ⁡ A cos ⁡ B − sin ⁡ B cos ⁡ A = 3 5 ( sin ⁡ A cos ⁡ B + sin ⁡ B cos ⁡ A ) ⇔ 2 sin ⁡ A cos ⁡ B = 8 sin ⁡ B cos ⁡ A ⇔ sin ⁡ A cos ⁡ B sin ⁡ B cos ⁡ A = 4 ⇔ tan ⁡ A tan ⁡ B = 4 \begin{aligned} a\cos B-b\cos A=\dfrac{3}{5}c & \Leftrightarrow 2R\sin A\cos B-2R\sin B\cos A=\dfrac{3}{5}2R\sin C \\ & \Leftrightarrow \sin A\cos B-\sin B\cos A=\dfrac{3}{5}\sin C \\ & \Leftrightarrow \sin A\cos B-\sin B\cos A=\dfrac{3}{5}\sin \left( A+B \right) \\ & \Leftrightarrow \sin A\cos B-\sin B\cos A=\dfrac{3}{5}\left( \sin A\cos B+\sin B\cos A \right) \\ & \Leftrightarrow 2\sin A\cos B=8\sin B\cos A \\ & \Leftrightarrow \dfrac{\sin A\cos B}{\sin B\cos A}=4 \\ & \Leftrightarrow \dfrac{\tan A}{\tan B}=\boxed{4} \\ \end{aligned}

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From the Projection Formula :

c c = a cos ⁡ B + b cos ⁡ A a\cos{B} + b\cos{A}

3 5 c \frac{3}{5} c = a cos ⁡ B − b cos ⁡ A a\cos{B} -b\cos{A}

Thus we get a cos ⁡ B a\cos{B} = 4 5 c \frac{4}{5} c and b cos ⁡ A b\cos{A} = 1 5 c \frac{1}{5} c

Thus

tan ⁡ A tan ⁡ B \frac{\tan{A}}{\tan{B}} = sin ⁡ A sin ⁡ B \frac{\sin{A}}{\sin{B}} . cos ⁡ B cos ⁡ A \frac{\cos{B}}{\cos{A}}

= a b \frac{ \cancel{a}}{\cancel{b}} . a cos ⁡ B b cos ⁡ A \frac{a\cos{B}}{b\cos{A}} . b a \frac{\cancel{b}}{\cancel{a}} [ From Law of Sines, sin ⁡ A sin ⁡ B \frac{\sin{A}}{\sin{B}} = a b \frac{a}{b} ]

= 4 c 5 c 5 \frac{\frac{4c}{5}}{\frac{c}{5}}

= 4 4

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Agent T
May 3, 2021

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David Vreken
May 4, 2021

By the law of cosines, cos ⁡ A = b 2 + c 2 − a 2 2 b c \cos A = \cfrac{b^2 + c^2 - a^2}{2bc} and cos ⁡ B = a 2 + c 2 − b 2 2 a c \cos B = \cfrac{a^2 + c^2 - b^2}{2ac} .

Substituting these into a cos ⁡ B − b cos ⁡ A = 3 5 c a \cos B - b \cos A = \cfrac{3}{5}c gives a ⋅ a 2 + c 2 − b 2 2 a c − b ⋅ b 2 + c 2 − a 2 2 b c = 3 5 c a \cdot \cfrac{a^2 + c^2 - b^2}{2ac} - b \cdot \cfrac{b^2 + c^2 - a^2}{2bc} = \cfrac{3}{5}c , which simplifies to a 2 − b 2 = 3 5 c 2 a^2 - b^2 = \cfrac{3}{5}c^2 .

Substituting a 2 − b 2 = 3 5 c 2 a^2 - b^2 = \cfrac{3}{5}c^2 back into cos ⁡ A \cos A and cos ⁡ B \cos B gives cos ⁡ A = c 2 − 3 5 c 2 2 b c = c 5 b \cos A = \cfrac{c^2 - \frac{3}{5}c^2}{2bc} = \cfrac{c}{5b} and cos ⁡ B = c 2 + 3 5 c 2 2 a c = 4 c 5 a \cos B = \cfrac{c^2 + \frac{3}{5}c^2}{2ac} = \cfrac{4c}{5a} , so cos ⁡ B cos ⁡ A = 4 c 5 a c 5 b = 4 b a \cfrac{\cos B}{\cos A} = \cfrac{\frac{4c}{5a}}{\frac{c}{5b}} = \cfrac{4b}{a} .

By the law of sines, sin ⁡ A sin ⁡ B = a b \cfrac{\sin A}{\sin B} = \cfrac{a}{b} .

Therefore, tan ⁡ A tan ⁡ B = sin ⁡ A sin ⁡ B ⋅ cos ⁡ B cos ⁡ A = a b ⋅ 4 b a = 4 \cfrac{\tan A}{\tan B} = \cfrac{\sin A}{\sin B} \cdot \cfrac{\cos B}{\cos A} = \cfrac{a}{b} \cdot \cfrac{4b}{a} = \boxed{4} .

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