Triangle Base D8E22C

1. **State the problem:** We have a triangle with base angles $55^\circ$ and $38^\circ$, and an altitude (height) of $108$ units from the top vertex to the base. We need to find the length of the base $x$. 2. **Identify the missing angle:** The sum of angles in a triangle is $180^\circ$. $$\text{Third angle} = 180^\circ - 55^\circ - 38^\circ = 87^\circ$$ 3. **Label the triangle:** Let the triangle be $\triangle ABC$ with $AB$ and $BC$ as the sides adjacent to the base $AC$. The altitude $h = 108$ is drawn from vertex $B$ perpendicular to base $AC$. 4. **Split the base into two segments:** Let the foot of the altitude be point $D$ on $AC$. Then $AD = x_1$ and $DC = x_2$, so $x = x_1 + x_2$. 5. **Use right triangles to find $x_1$ and $x_2$:** In right triangle $ABD$, angle at $A$ is $55^\circ$. $$\tan 55^\circ = \frac{h}{x_1} \implies x_1 = \frac{h}{\tan 55^\circ} = \frac{108}{\tan 55^\circ}$$ In right triangle $BCD$, angle at $C$ is $38^\circ$. $$\tan 38^\circ = \frac{h}{x_2} \implies x_2 = \frac{h}{\tan 38^\circ} = \frac{108}{\tan 38^\circ}$$ 6. **Calculate $x_1$ and $x_2$:** $$x_1 = \frac{108}{\tan 55^\circ} \approx \frac{108}{1.4281} \approx 75.64$$ $$x_2 = \frac{108}{\tan 38^\circ} \approx \frac{108}{0.7813} \approx 138.28$$ 7. **Find total base length $x$:** $$x = x_1 + x_2 = 75.64 + 138.28 = 213.92$$ **Final answer:** $$\boxed{x \approx 213.92}$$