Question

Solve for each variable. Attached is a picture of the equation.

Answer 1

Write the equation

`y\text{ = tan}\theta\text{ . x - }(g)/(2u^2.cos^2\theta).x^2`

1)

`\begin{gathered} \text{Solve for }\theta \\ \sec ^2\theta\text{ = }(1)/(\cos^2\theta) \\ 1+tan^2\theta=sec^2\theta \\ y\text{ = xtan}\theta\text{ - }(g\sec^2\theta)/(2u^2).x^2 \\ y\text{ = xtan}\theta\text{ - }(gx^2)/(2u^2)\sec e^2\theta \\ y\text{ = xtan}\theta\text{ - }(gx^2)/(2u^2)(1+tan^2\theta) \\ y\text{ = xtan}\theta\text{ - }\frac{gx^2}{2u^2\text{ }}\text{ - }(gx^2)/(2u^2)\tan ^2\theta \\ (gx^2)/(2u^2)\tan ^2\theta\text{ - xtan}\theta\text{ + y + }(gx^2)/(2u^2)\text{ = 0} \\ \text{Let tan}\theta\text{ = m} \\ (gx^2)/(2u^2)m^2\text{ - xm + y + }(gx^2)/(2u^2)\text{ = 0} \\ m\text{ = }\frac{x\text{ }\pm\text{ }\sqrt[]{x^2\text{ - 4(}(gx^2)/(2u^2))(y+(gx^2)/(2u^2))}}{(gx)/(u^2)} \\ m=sec^2\theta \\ \sec ^2\theta\text{ = }\frac{x\text{ }\pm\sqrt[]{x^2\text{ - 4(}(gx^2)/(2u^2))(y\text{ + }(gx^2)/(2u^2))}}{(gx)/(u^2)} \\ \\ \sec \theta\text{ = }\sqrt[]{\frac{x\text{ }\pm\sqrt[]{x^2\text{ - 4(}(gx^2)/(2u^2))(y+(gx^2)/(2u^2))}}{(gx^2)/(u^2)}} \\ \theta=sec^(-1)(\sqrt[]{\frac{x\pm\sqrt[]{x^2\text{ - 4(}(gx^2)/(2u^2))(y\text{ + }(gx^2)/(2u^2))}}{(gx^2)/(u^2)}}) \end{gathered}`

2) Make x subject of the relation

`\begin{gathered} y\text{ = xtan}\theta\text{ - }(g)/(2u^2\cos ^2\theta)x^2 \\ (g)/(2u^2\cos ^2\theta)x^2\text{ - xtan}\theta\text{ + y = 0} \\ x\text{ = }\frac{\tan \theta\pm\sqrt[]{\tan^2\theta-4((g)/(2u^2\cos^2\theta))y}}{2((g)/(2u^2\cos ^2\theta))} \\ x=\text{ }\frac{\tan \theta\text{ }\pm\text{ }\sqrt[]{\tan^2\theta\text{ - }(2gy)/(u^2\cos^2\theta)}}{(g)/(u^2\cos ^2\theta)} \end{gathered}`