1. **Problem 1:** Given triangle PQR with sides $q=12$ cm, $r=16$ cm, and angle $P=54^\circ$, find the third side or other unknowns if needed. 2. **Problem 2:** Given triangle PQR with sides $q=3.25$ m, $r=4.42$ m, and angle $P=105^\circ$, find the third side or other unknowns if needed. 3. **Problem 3:** Given triangle XYZ with sides $x=100$ cm, $y=8.0$ cm, and $z=7.0$ cm, analyze or find angles or other properties. 4. **Problem 4:** Given triangle XYZ with sides $x=21$ mm, $y=34$ mm, and $z=42$ mm, analyze or find angles or other properties. --- ### Step 1: Understanding the Law of Cosines The Law of Cosines relates the lengths of sides of a triangle to the cosine of one of its angles: $$c^2 = a^2 + b^2 - 2ab\cos(C)$$ where $c$ is the side opposite angle $C$, and $a$, $b$ are the other two sides. This formula helps find the unknown side or angle in a triangle when two sides and the included angle are known. --- ### Step 2: Solve Problems 1 and 2 using Law of Cosines For triangle PQR, let side $p$ be opposite angle $P$. **Problem 1:** $$p^2 = q^2 + r^2 - 2qr\cos(P)$$ $$p^2 = 12^2 + 16^2 - 2 \times 12 \times 16 \times \cos(54^\circ)$$ Calculate: $$p^2 = 144 + 256 - 384 \times \cos(54^\circ)$$ $$\cos(54^\circ) \approx 0.5878$$ $$p^2 = 400 - 384 \times 0.5878 = 400 - 225.7152 = 174.2848$$ $$p = \sqrt{174.2848} \approx 13.2 \text{ cm}$$ **Problem 2:** $$p^2 = 3.25^2 + 4.42^2 - 2 \times 3.25 \times 4.42 \times \cos(105^\circ)$$ Calculate: $$p^2 = 10.5625 + 19.5364 - 28.715 \times \cos(105^\circ)$$ $$\cos(105^\circ) \approx -0.2588$$ $$p^2 = 30.0989 - 28.715 \times (-0.2588) = 30.0989 + 7.433 = 37.5319$$ $$p = \sqrt{37.5319} \approx 6.13 \text{ m}$$ --- ### Step 3: Analyze Problems 3 and 4 using Triangle Inequality and Angles For triangles XYZ with sides $x$, $y$, $z$, check if the triangle is valid using the triangle inequality: $$x + y > z, \quad y + z > x, \quad z + x > y$$ **Problem 3:** $$100 + 8 > 7 \Rightarrow 108 > 7 \text{ (True)}$$ $$8 + 7 > 100 \Rightarrow 15 > 100 \text{ (False)}$$ Since one inequality fails, triangle with sides 100 cm, 8 cm, 7 cm is not valid. **Problem 4:** $$21 + 34 > 42 \Rightarrow 55 > 42 \text{ (True)}$$ $$34 + 42 > 21 \Rightarrow 76 > 21 \text{ (True)}$$ $$42 + 21 > 34 \Rightarrow 63 > 34 \text{ (True)}$$ All inequalities hold, so triangle is valid. To find angles, use Law of Cosines, for example angle $X$ opposite side $x$: $$\cos(X) = \frac{y^2 + z^2 - x^2}{2yz}$$ Calculate angle $X$: $$\cos(X) = \frac{34^2 + 42^2 - 21^2}{2 \times 34 \times 42} = \frac{1156 + 1764 - 441}{2856} = \frac{2479}{2856} \approx 0.868$$ $$X = \cos^{-1}(0.868) \approx 30.2^\circ$$ Similarly, angles $Y$ and $Z$ can be found using the same formula. --- ### Final answers: 1. Side $p \approx 13.2$ cm 2. Side $p \approx 6.13$ m 3. Triangle invalid (does not satisfy triangle inequality) 4. Triangle valid, angle $X \approx 30.2^\circ$ (others can be found similarly)