Triangles triangles triangles

Geometry Level 4

In a triangle A B C ABC , if a 2 tan ⁡ B = b 2 tan ⁡ A a^2\tan B= b^2\tan A , what triangle would △ A B C \triangle ABC be?

A special right angle triangle A isosceles triangle A isosceles triangle or right-angle triangle A right-angle triangle

Xi Guan by xi guan , 20, Taiwan

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1 solution

Zico Quintina
May 20, 2018

In the right triangle on the left, we have

sin ⁡ A = a c = cos ⁡ B ( 1 ) sin ⁡ B = b c = cos ⁡ A ( 2 ) a sin ⁡ A = b sin ⁡ B [Sine Law] a ⋅ a sin ⁡ A ⋅ sin ⁡ A = b ⋅ b sin ⁡ B ⋅ sin ⁡ B a 2 sin ⁡ A ⋅ cos ⁡ B = b 2 sin ⁡ B ⋅ cos ⁡ A [Substituting with (1) and (2)] a 2 sin ⁡ B cos ⁡ B = b 2 sin ⁡ A cos ⁡ A [Multiplying by sin ⁡ A sin ⁡ B ] a 2 tan ⁡ B = b 2 tan ⁡ A \begin{aligned} \sin A = &\dfrac{a}{c} = \cos B \qquad (1) \\ \\ \sin B = &\dfrac{b}{c} = \cos A \qquad (2) \\ \\ \dfrac{a}{\sin A} &= \dfrac{b}{\sin B} \qquad \qquad \quad \ \small \text{[Sine Law]} \\ \\ \dfrac{a \cdot a}{\sin A \cdot \sin A} &= \dfrac{b \cdot b}{\sin B \cdot \sin B} \\ \\ \dfrac{a^2}{\sin A \cdot \cos B} &= \dfrac{b^2}{\sin B \cdot \cos A} \qquad \, \small \text{[Substituting with (1) and (2)]} \\ \\ \dfrac{a^2 \sin B}{\cos B} &= \dfrac{b^2 \sin A}{\cos A} \qquad \qquad \small \text{[Multiplying by } \sin A \sin B] \\ \\ a^2 \tan B &= b^2 \tan A \end{aligned} In the isosceles triangle on the right, we have

a = b tan ⁡ B = tan ⁡ A a 2 tan ⁡ B = b 2 tan ⁡ A \begin{aligned} a &= b \\ \\ \tan B &= \tan A \\ \\ a^2 \tan B &= b^2 \tan A \end{aligned}

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