1. **Stating the problem:** We have two right triangles each with an angle of 29°. - In the first triangle, the side adjacent to the 29° angle is 15, and the hypotenuse is $x$. - In the second triangle, the side opposite the 29° angle is 15, and the adjacent side is $x$. We want to find $x$ in each case. 2. **First triangle:** Given adjacent side and hypotenuse. We know the cosine of an angle in a right triangle is: $$\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$$ Substitute $\theta = 29^\circ$, adjacent $= 15$, hypotenuse $= x$: $$\cos(29^\circ) = \frac{15}{x} \implies x = \frac{15}{\cos(29^\circ)}$$ Calculate the value: $$\cos(29^\circ) \approx 0.8746$$ Therefore, $$x \approx \frac{15}{0.8746} \approx 17.16$$ 3. **Second triangle:** Given opposite side and adjacent side. We know the tangent of an angle in a right triangle is: $$\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}$$ Substitute $\theta = 29^\circ$, opposite $= 15$, adjacent $= x$: $$\tan(29^\circ) = \frac{15}{x} \implies x = \frac{15}{\tan(29^\circ)}$$ Calculate the value: $$\tan(29^\circ) \approx 0.5543$$ Therefore, $$x \approx \frac{15}{0.5543} \approx 27.06$$ **Final answers:** - For the first triangle: $x \approx 17.16$ - For the second triangle: $x \approx 27.06$