[Trigonometric ratios in a right triangle] [Trigonometric equations

Isosceles triangles $ABC (AB = BC)$ and $A_{1}B_{1}C_{1} (A_{1}B_{1} = B_{1}C_{1})$ are similar, and $AC: A_{1}C_{1} = 5: \sqrt{3}$.

Vertices $A_{1}$ and $B_{1}$ are located on sides $AC$ and $BC$, respectively, and vertex $C_{1}$ is on the extension of side $AB$ beyond point $B$, such that $A_{1}B_{1} \perp BC$. Find the angle $B$.

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Prove that $C_{1} A_{1} \perp A C$ and formulate a trigonometric equation with respect to the angle at the base of the isosceles triangle $A B C$.

## Solution

Let $A_{1} B_{1}=B_{1} C_{1}=1, \angle C=\gamma$. Then $A B=\frac{5}{\sqrt{3}}, \angle C_{1} A_{1} C=\angle C_{1} A_{1} B_{1}+\angle B_{1} A_{1} C=\gamma+\left(90^{\circ}-\gamma\right)=90^{\circ}, A_{1} C$ $=1 / \sin \gamma, C_{1} A_{1}=2 \cos \gamma, A A_{1}=C_{1} A_{1} \operatorname{ctg} \gamma=\frac{\cos ^{2} \gamma}{\sin \gamma}, A C=\frac{10}{\sqrt{3}} \cos \gamma$.

Since $A C=A A_{1}+A_{1} C$, then

$\frac{10 \cos \gamma}{\sqrt{3}}=\frac{\cos ^{2} \gamma}{\sin \gamma}+\frac{1}{\sin \gamma} \Leftrightarrow \frac{10 \cos \gamma \sin \gamma}{\sqrt{3}}=2 \cos ^{2} \gamma+\sin ^{2} \gamma+\cos ^{2} \gamma \Leftrightarrow$ $\sqrt{3} \operatorname{tg}^{2} \gamma-10 \operatorname{tg} \gamma+3 \sqrt{3}=0$

From this, $\operatorname{tg} \gamma=\frac{1}{\sqrt{3}}$ or $\operatorname{tg} \gamma=3 \sqrt{3}$.

The second solution does not satisfy the condition: point $C_{1}$ lies on the extension of segment $A B$ beyond point $B$, so $A B \sin \gamma<A_{1} C_{1}$, that is, $\operatorname{tg} \gamma<\frac{2 \sqrt{3}}{5}$. Therefore, $\gamma=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-19.jpg?height=303&width=954&top_left_y=0&top_left_x=555)

## Answer

$120^{\circ}$.