An isosceles triangle has equal sides a = 4 cm

An isosceles triangle has equal sides $a = 4\ \text{cm}$ and included angle $\alpha = 60^{\circ}$. By how much must $\alpha$ increase so that the area increases by $0.16\ \text{cm}^{2}$?

Step 1: The area as a function of the angle

For two sides $a$ with included angle $\alpha$, \[S = \frac12 a^{2}\sin\alpha = \frac12\cdot 16\cdot\sin\alpha = 8\sin\alpha \quad (\text{cm}^{2}).\]

Only $\alpha$ varies; the sides are fixed.

Step 2: Differentiate

\[ \frac{dS}{d\alpha} = 8\cos\alpha, \qquad \Delta S \approx dS = 8\cos\alpha\,\Delta\alpha. \]

Step 3: Solve for $\Delta\alpha$

At $\alpha = 60^{\circ}$ we have $\cos\alpha = \tfrac12$, so $dS/d\alpha = 4\ \text{cm}^{2}$ per radian: $\Delta\alpha \approx \frac{\Delta S}{8\cos\alpha} = \frac{0.16}{4} = 0.04\ \text{rad}$.

Step 4: Convert to degrees

\[ 0.04\ \text{rad} = 0.04\cdot\frac{180^{\circ}}{\pi} \approx 2.2918^{\circ} \approx 2^{\circ}17.5'. \]

Step 5: Verification

\[S(60^{\circ}) = 8\sin 60^{\circ} = 4\sqrt3 \approx 6.92820\ \text{cm}^{2}, \qquad S_{\text{target}} = 7.08820\ \text{cm}^{2}\]

\[\sin\alpha_{2} = \frac{7.08820}{8} = 0.8860254 \;\Longrightarrow\; \alpha_{2} \approx 62.378^{\circ}\]

\[\Delta\alpha_{\text{exact}} \approx 2.378^{\circ} = 0.04150\ \text{rad versus } 0.04\ \text{rad} \;\;\checkmark \text{ (estimate low by } 3.6\%)\]