dl1/dx = x / √(m² + x²) differentiate l2. dl2/dx = (x - d) / √(n² + (d - x)²) solve the equation dl1/dx +…

dl1/dx = x / √(m² + x²) differentiate l2. dl2/dx = (x - d) / √(n² + (d - x)²) solve the equation dl1/dx + dl2/dx = 0 for x in the interval 0,d. x = md / (m + n) use the expression for x to find θ1 in terms of m, n, and d. θ1 = tan⁻¹( )

dl1/dx = x / √(m² + x²) differentiate l2. dl2/dx = (x - d) / √(n² + (d - x)²) solve the equation dl1/dx + dl2/dx = 0 for x in the interval 0,d. x = md / (m + n) use the expression for x to find θ1 in terms of m, n, and d. θ1 = tan⁻¹( )

Answer

Explanation:

Step1: Recall the tangent - angle relationship

We know that if we consider a right - triangle, $\tan\theta_1=\frac{m}{x}$. We have found that $x = \frac{md}{m + n}$.

Step2: Substitute the value of (x) into the tangent formula

Substitute (x=\frac{md}{m + n}) into (\tan\theta_1=\frac{m}{x}). Then (\tan\theta_1=\frac{m}{\frac{md}{m + n}}).

Step3: Simplify the expression

(\tan\theta_1=\frac{m + n}{d}). So, (\theta_1=\tan^{-1}\left(\frac{m + n}{d}\right)).

Answer:

(\frac{m + n}{d})