1. **Problem a:** Given right triangle G T M with right angle at G, side G T = 33.4 m (vertical), diagonal T M = 48.2 m, and angle $x^\circ$ at M between G M and T M. Find $x$. 2. Use the cosine formula for angle $x$ in right triangle: $$\cos(x) = \frac{\text{adjacent side}}{\text{hypotenuse}}$$ Here, adjacent side to angle $x$ is G M (horizontal side), hypotenuse is T M = 48.2 m. 3. Find G M using Pythagoras theorem: $$G M = \sqrt{(T M)^2 - (G T)^2} = \sqrt{48.2^2 - 33.4^2}$$ Calculate: $$48.2^2 = 2323.24, \quad 33.4^2 = 1115.56$$ $$G M = \sqrt{2323.24 - 1115.56} = \sqrt{1207.68} \approx 34.76$$ 4. Calculate $\cos(x)$: $$\cos(x) = \frac{34.76}{48.2} \approx 0.721$$ 5. Find $x$ by inverse cosine: $$x = \cos^{-1}(0.721) \approx 43.8^\circ$$ 6. **Problem b:** Right triangle with angle 58° at left vertex, right angle at bottom-right vertex, bottom side = 40 yd, top side = $x$. Find $x$. 7. Use tangent formula: $$\tan(58^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{40}$$ 8. Solve for $x$: $$x = 40 \times \tan(58^\circ)$$ Calculate: $$\tan(58^\circ) \approx 1.6003$$ $$x \approx 40 \times 1.6003 = 64.0$$ **Final answers:** a. $x \approx 43.8^\circ$ b. $x \approx 64.0$ yards