Triangle Area Sides 744235

1. **State the problem:** We have triangle ABC with sides $AB = 15$ cm, $AC = 25$ cm, angle $BAC = \theta^\circ$, and area $= 100$ cm². We need to find (a) $\sin \theta$ and (b) the length of side $BC$ given $\theta > 90^\circ$. 2. **Formula for area of a triangle using two sides and included angle:** $$\text{Area} = \frac{1}{2} \times AB \times AC \times \sin \theta$$ 3. **Calculate $\sin \theta$:** Given area $= 100$, substitute values: $$100 = \frac{1}{2} \times 15 \times 25 \times \sin \theta$$ Simplify: $$100 = \frac{1}{2} \times 375 \times \sin \theta = 187.5 \sin \theta$$ Divide both sides by 187.5: $$\sin \theta = \frac{100}{187.5}$$ Show cancellation: $$\sin \theta = \frac{\cancel{100}}{\cancel{187.5}} = \frac{8}{15}$$ Numerically: $$\sin \theta = 0.5333...$$ 4. **Interpretation of $\sin \theta$ given $\theta > 90^\circ$:** Since $\sin \theta = 0.5333$ and $\theta > 90^\circ$, $\theta$ is in the second quadrant where sine is positive. 5. **Find $\theta$ (for understanding):** $$\theta = 180^\circ - \arcsin(0.5333)$$ But exact value is not required here. 6. **Find length of BC using Law of Cosines:** Law of Cosines: $$BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos \theta$$ We know $AB = 15$, $AC = 25$, and $\sin \theta = 8/15$. Find $\cos \theta$ using Pythagorean identity: $$\cos \theta = -\sqrt{1 - \sin^2 \theta}$$ Negative because $\theta > 90^\circ$ (cosine negative in second quadrant). Calculate: $$\sin^2 \theta = \left(\frac{8}{15}\right)^2 = \frac{64}{225}$$ $$\cos \theta = -\sqrt{1 - \frac{64}{225}} = -\sqrt{\frac{225 - 64}{225}} = -\sqrt{\frac{161}{225}} = -\frac{\sqrt{161}}{15}$$ 7. **Calculate $BC^2$:** $$BC^2 = 15^2 + 25^2 - 2 \times 15 \times 25 \times \left(-\frac{\sqrt{161}}{15}\right)$$ Simplify: $$BC^2 = 225 + 625 + 2 \times 25 \times \sqrt{161}$$ $$BC^2 = 850 + 50 \sqrt{161}$$ 8. **Calculate numerical value:** $$\sqrt{161} \approx 12.6886$$ $$BC^2 \approx 850 + 50 \times 12.6886 = 850 + 634.43 = 1484.43$$ 9. **Find $BC$:** $$BC = \sqrt{1484.43} \approx 38.53$$ 10. **Final answers:** - (a) $\sin \theta = \frac{8}{15}$ - (b) $BC \approx 38.5$ cm (to 3 significant figures)