1. **State the problem:** We are given a triangle with sides $a=19$ cm, $c=21$ cm, and the segment $s_{156}=15$ cm (likely a side or segment related to the triangle). We want to solve the triangle, meaning find all unknown sides and angles. 2. **Identify the formula:** The area $S$ of the triangle is given by Heron's formula: $$S = \sqrt{s(s-a)(s-b)(s-c)}$$ where $s$ is the semi-perimeter: $$s = \frac{a+b+c}{2}$$ 3. **Given:** - $a=19$ cm - $c=21$ cm - $s_{156}=15$ cm (interpreted as side $b=15$ cm) - Semi-perimeter $s = 27.5$ cm (given) 4. **Calculate the area using Heron's formula:** $$S = \sqrt{27.5(27.5-19)(27.5-15)(27.5-21)}$$ Calculate each term: $$27.5 - 19 = 8.5$$ $$27.5 - 15 = 12.5$$ $$27.5 - 21 = 6.5$$ So, $$S = \sqrt{27.5 \times 8.5 \times 12.5 \times 6.5}$$ Calculate the product inside the root: $$27.5 \times 8.5 = 233.75$$ $$12.5 \times 6.5 = 81.25$$ $$233.75 \times 81.25 = 18984.375$$ Therefore, $$S = \sqrt{18984.375} \approx 137.8 \text{ cm}^2$$ 5. **Find the angles using the Law of Cosines:** For angle $B$ opposite side $b=15$ cm: $$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{19^2 + 21^2 - 15^2}{2 \times 19 \times 21}$$ Calculate numerator: $$19^2 = 361, \quad 21^2 = 441, \quad 15^2 = 225$$ $$361 + 441 - 225 = 577$$ Calculate denominator: $$2 \times 19 \times 21 = 798$$ So, $$\cos B = \frac{577}{798} \approx 0.7231$$ Angle $B$: $$B = \arccos(0.7231) \approx 43.8^\circ$$ 6. **Find angle $A$ using Law of Cosines:** $$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{15^2 + 21^2 - 19^2}{2 \times 15 \times 21}$$ Calculate numerator: $$225 + 441 - 361 = 305$$ Calculate denominator: $$2 \times 15 \times 21 = 630$$ So, $$\cos A = \frac{305}{630} \approx 0.4841$$ Angle $A$: $$A = \arccos(0.4841) \approx 61.1^\circ$$ 7. **Find angle $C$:** Sum of angles in triangle is $180^\circ$: $$C = 180^\circ - A - B = 180^\circ - 61.1^\circ - 43.8^\circ = 75.1^\circ$$ **Final answers:** - Area $S \approx 137.8$ cm$^2$ - Angles: $A \approx 61.1^\circ$, $B \approx 43.8^\circ$, $C \approx 75.1^\circ$