1. The problem involves a triangle with sides 90 cm, 130 cm, and a side labeled $e$. Angles opposite these sides are $f$ and $g$. We want to find equations for $e$, $f$, and $g$ using SOHCAHTOA. 2. SOHCAHTOA relates sides and angles in right triangles: - Sine: $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$ - Cosine: $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$ - Tangent: $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$ 3. Identify sides relative to angles $f$ and $g$: - For angle $f$, opposite side is $e$, adjacent side is 90 cm, hypotenuse is 130 cm. - For angle $g$, opposite side is 90 cm, adjacent side is $e$, hypotenuse is 130 cm. 4. Write equations: - Using sine for angle $f$: $$\sin(f) = \frac{e}{130} \implies e = 130 \sin(f)$$ - Using cosine for angle $f$: $$\cos(f) = \frac{90}{130}$$ (can find $f$ from this) - Using tangent for angle $g$: $$\tan(g) = \frac{90}{e} \implies g = \tan^{-1}\left(\frac{90}{e}\right)$$ 5. Summary of equations: $$e = 130 \sin(f)$$ $$f = \cos^{-1}\left(\frac{90}{130}\right)$$ $$g = \tan^{-1}\left(\frac{90}{e}\right)$$ These use SOHCAHTOA to relate sides and angles in the triangle.