Triangle Solving – Problem 17: Multiple choice (select all that apply)

Multiple choice (select all that apply). Which of the following statements are true?

A. In $\triangle ABC$, if $A>B$ then $\sin A>\sin B$ B. In an acute $\triangle ABC$, the inequality $\sin A>\cos B$ always holds C. In $\triangle ABC$, if $a\cos A=b\cos B$, then $\triangle ABC$ must be an isosceles right triangle D. In $\triangle ABC$, if $B=60^{\circ}$ and $b^{2}=ac$, then $\triangle ABC$ must be equilateral

Step 1. A: If $A>B$, then $a>b$. By the Law of Sines $\frac{a}{\sin A}=\frac{b}{\sin B}$, we have $\frac{\sin A}{\sin B}=\frac{a}{b}>1$, so $\sin A>\sin B$. Thus A is true.

Step 2. B: In an acute triangle, $A+B>\frac{\pi}{2}$, so $A>\frac{\pi}{2}-B$. Since $\sin x$ is increasing on $(0,\frac{\pi}{2})$, $\sin A>\sin\left(\frac{\pi}{2}-B\right)=\cos B$. Thus B is true.

Step 3. C: If $a\cos A=b\cos B$, then using $\frac{a}{b}=\frac{\sin A}{\sin B}$ we get $\sin A\cos A=\sin B\cos B$, i.e. $\sin 2A=\sin 2B$. Hence $A=B$ or $A+B=\frac{\pi}{2}$, so the triangle is isosceles or right, not necessarily both. Thus C is false.

Step 4. D: If $B=60^{\circ}$ and $b^{2}=ac$, then by the Law of Cosines $$b^{2}=a^{2}+c^{2}-2ac\cos B=a^{2}+c^{2}-ac.$$ So $ac=a^{2}+c^{2}-ac\Rightarrow (a-c)^{2}=0\Rightarrow a=c$. Then $A=C$ and $A+C=120^{\circ}$, giving $A=C=60^{\circ}$. Hence the triangle is equilateral, so D is true.

Step 5. Therefore, the correct choices are A, B, and D.

ABD