1. **Problem statement:** We have an isosceles triangle ABC with sides AB = AC, angle $A = 48^\circ$, and a circle inscribed in the triangle. The circle's radius (inradius) is 11 cm. We are asked to find the length of AC, correct to 1 decimal place. 2. **Label and known values:** - Let $AB = AC = x$ (the equal sides to find). - Base $BC$ is unknown. - Angle $A = 48^\circ$. - Inradius $r = 11$ cm. 3. **Key formulas:** - Area $\triangle ABC = A = r \times s$ where $s$ is the semiperimeter. - Semiperimeter $s = \frac{a+b+c}{2}$. - Since $AB = AC = x$ and $BC = b$, $s = \frac{x + x + b}{2} = \frac{2x + b}{2} = x + \frac{b}{2}$. 4. **Expressing base $BC$ in terms of $x$ and angle $A$:** - Drop perpendiculars from $A$ to $BC$, creating two right triangles. - Since $AB=AC=x$ and $ABC$ is isosceles, $BC$ can be represented using Law of Cosines or sine: Using Law of Cosines on $\triangle ABC$, $$ b^2 = x^2 + x^2 - 2x^2 \cos 48^\circ = 2x^2 (1 - \cos 48^\circ) $$ so, $$ b = x \sqrt{2(1-\cos 48^\circ)} $$ 5. **Area in terms of $x$:** Area $= \frac{1}{2} (BC) (\text{height from } A)$. Height $h = x \sin 24^\circ$ since the altitude bisects the vertex angle 48° into two angles of 24° each (isosceles triangle property). Base $BC = 2x \sin 24^\circ$ (since the base segments are equal and adjacent to angle 24°, by sine rule in half triangle). Area: $$ A = \frac{1}{2} \times BC \times h = \frac{1}{2} \times 2x \sin 24^\circ \times x \sin 24^\circ = x^2 \sin^2 24^\circ $$ 6. **Using the inradius formula:** $$ A = r \times s = 11 \times \left(x + \frac{b}{2}\right) $$ From Step 4, $$ b = 2x \sin 24^\circ $$ thus, $$ s = x + x \sin 24^\circ = x(1 + \sin 24^\circ) $$ 7. **Equate the two expressions for area:** $$ x^2 \sin^2 24^\circ = 11 \times x (1 + \sin 24^\circ) $$ Divide both sides by $x$ (nonzero): $$ x \sin^2 24^\circ = 11 (1 + \sin 24^\circ) $$ 8. **Solve for $x$:** $$ x = \frac{11(1 + \sin 24^\circ)}{\sin^2 24^\circ} $$ Calculate values: - $\sin 24^\circ \approx 0.4067$ - $\sin^2 24^\circ \approx 0.1654$ - $1 + 0.4067 = 1.4067$ Therefore, $$ x \approx \frac{11 \times 1.4067}{0.1654} = \frac{15.4737}{0.1654} \approx 93.55 $$ 9. **Final answer:** Length of $AC$ is approximately $93.6$ cm (rounded to 1 decimal place).