Precalculus II Unit Test Revie

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### Solving Oblique Triangles (Non-Right) #### Key Concepts for Oblique Triangles - **Oblique triangles** are triangles that do not contain a right angle. - **Solving a triangle** means finding the measures of all three sides and all three angles. - **Notation:** Angles are typically denoted by capital letters ($A, B, C$) and the sides opposite them by corresponding lowercase letters ($a, b, c$). ### Law of Sines #### Formula $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ #### When to Use Law of Sines Use the Law of Sines when you know: 1. **Angle-Side-Angle (ASA):** Two angles and the included side. 2. **Angle-Angle-Side (AAS):** Two angles and a non-included side. 3. **Side-Side-Angle (SSA):** Two sides and a non-included angle (Ambiguous Case - might have 0, 1, or 2 solutions). #### Ambiguous Case (SSA) Flowchart Given sides $a, b$ and angle $A$: 1. **Calculate $h = b \sin A$.** (height of the triangle relative to side $b$ and angle $A$) 2. **Compare $a$ with $h$ and $b$:** * **If $a ### Law of Cosines #### Formulas $$a^2 = b^2 + c^2 - 2bc \cos A$$ $$b^2 = a^2 + c^2 - 2ac \cos B$$ $$c^2 = a^2 + b^2 - 2ab \cos C$$ (To find an angle, rearrange: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$, etc.) #### When to Use Law of Cosines Use the Law of Cosines when you know: 1. **Side-Angle-Side (SAS):** Two sides and the included angle. 2. **Side-Side-Side (SSS):** All three sides. #### Example Problem (SAS, then Area) Given: $A = 50^\circ$, $b = 15$, $c = 30$. 1. **Find $a$ using Law of Cosines:** $$a^2 = 15^2 + 30^2 - 2(15)(30) \cos 50^\circ \approx 546.48 \implies a \approx 23.38$$ 2. **Calculate Area using SAS formula:** $$Area = \frac{1}{2}bc \sin A = \frac{1}{2}(15)(30) \sin 50^\circ$$ $$Area = 225 \sin 50^\circ \approx 225(0.7660) \approx 172.35 \text{ units}^2$$ ### Area of Oblique Triangles #### Formulas 1. **SAS (Side-Angle-Side):** $$Area = \frac{1}{2}bc \sin A = \frac{1}{2}ac \sin B = \frac{1}{2}ab \sin C$$ (Use when you know two sides and the included angle.) 2. **SSS (Side-Side-Side) - Heron's Formula:** $$Area = \sqrt{s(s-a)(s-b)(s-c)}$$ where $s = \frac{a+b+c}{2}$ (the semi-perimeter). #### Example Problem (Heron's Formula) Given: $a = 55$ m, $b = 85$ m, $c = 100$ m. 1. **Calculate semi-perimeter $s$:** $s = \frac{55 + 85 + 100}{2} = 120$. 2. **Apply Heron's Formula:** $$Area = \sqrt{120(120-55)(120-85)(120-100)} = \sqrt{120(65)(35)(20)} \approx 2336.67 \text{ m}^2$$ --- ### Vectors #### Vector Representation - **Rectangular Coordinates (Component Form):** $\vec{v} = \langle a, b \rangle$ - **Linear Combination:** $\vec{v} = a\mathbf{i} + b\mathbf{j}$, where $\mathbf{i} = \langle 1, 0 \rangle$ and $\mathbf{j} = \langle 0, 1 \rangle$. #### Finding Component Form from Initial and Terminal Points If a vector has initial point $P_1(x_1, y_1)$ and terminal point $P_2(x_2, y_2)$, its component form is: $$\vec{v} = \langle x_2 - x_1, y_2 - y_1 \rangle$$ #### Magnitude of a Vector For $\vec{v} = \langle a, b \rangle$ or $\vec{v} = a\mathbf{i} + b\mathbf{j}$: $$||\vec{v}|| = \sqrt{a^2 + b^2}$$ #### Direction of a Vector For $\vec{v} = \langle a, b \rangle$: The direction angle $\theta$ is found using $\tan \theta = \frac{b}{a}$. * **Quadrant Check:** Adjust $\theta$ based on the quadrant of $\langle a, b \rangle$. * If $a > 0, b > 0$: $\theta = \arctan(\frac{b}{a})$ (Quadrant I) * If $a 0$: $\theta = \arctan(\frac{b}{a}) + 180^\circ$ (Quadrant II) * If $a 0, b ### Vector Operations #### Vector Addition/Subtraction If $\vec{u} = \langle u_1, u_2 \rangle$ and $\vec{v} = \langle v_1, v_2 \rangle$: - **Sum:** $\vec{u} + \vec{v} = \langle u_1 + v_1, u_2 + v_2 \rangle$ - **Difference:** $\vec{u} - \vec{v} = \langle u_1 - v_1, u_2 - v_2 \rangle$ #### Scalar Multiplication If $k$ is a scalar and $\vec{v} = \langle a, b \rangle$, then $k\vec{v} = \langle ka, kb \rangle$. #### Dot Product For $\vec{u} = \langle u_1, u_2 \rangle$ and $\vec{v} = \langle v_1, v_2 \rangle$: $$\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2$$ The result is a scalar. #### Angle Between Two Vectors The angle $\theta$ between two non-zero vectors $\vec{u}$ and $\vec{v}$ is given by: $$\cos \theta = \frac{\vec{u} \cdot \vec{v}}{||\vec{u}|| \, ||\vec{v}||}$$ $$\theta = \arccos\left(\frac{\vec{u} \cdot \vec{v}}{||\vec{u}|| \, ||\vec{v}||}\right)$$ #### Example Problem (Vector Operations with i, j) Given: $\vec{u} = 2\mathbf{i} + 2\mathbf{j}$ and $\vec{v} = -3\mathbf{i} + 4\mathbf{j}$. 1. **Sum:** $\vec{u} + \vec{v} = (2-3)\mathbf{i} + (2+4)\mathbf{j} = -\mathbf{i} + 6\mathbf{j}$ 2. **Difference:** $\vec{u} - \vec{v} = (2-(-3))\mathbf{i} + (2-4)\mathbf{j} = 5\mathbf{i} - 2\mathbf{j}$ 3. **Scalar Multiple:** $4\vec{u} = 4(2\mathbf{i} + 2\mathbf{j}) = 8\mathbf{i} + 8\mathbf{j}$ 4. **Dot Product:** $\vec{u} \cdot \vec{v} = (2)(-3) + (2)(4) = -6 + 8 = 2$ ### Force Problems (Conceptual) - **Represent forces as vectors:** Each force has a magnitude and a direction. - **Resolution into components:** Resolve forces into horizontal and vertical components. * For a force $\vec{F}$ with magnitude $||\vec{F}||$ at angle $\theta$ (from positive x-axis): $F_x = ||\vec{F}|| \cos \theta$ $F_y = ||\vec{F}|| \sin \theta$ * **Tip:** If direction is given as "x degrees in x direction", this usually means $x$ degrees from the positive x-axis. If it's a compass direction (North, East, etc.), convert to standard angle. - **Net Force:** The resultant force is the vector sum of all individual force vectors. * $\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 + ...$ * Calculate components of net force: $\vec{F}_{net} = \langle \sum F_{x}, \sum F_{y} \rangle$. - **Magnitude of Net Force:** $||\vec{F}_{net}|| = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$. - **Direction of Net Force:** $\theta_{net} = \arctan\left(\frac{\sum F_y}{\sum F_x}\right)$ (remember to adjust for quadrant). - **Equilibrium:** If an object is in equilibrium (not accelerating), the net force is zero: $\vec{F}_{net} = \langle 0, 0 \rangle$. #### Example Problem (Force with x degrees in x direction) Two forces act on an object. $\vec{F}_1$: 10 N at $30^\circ$ (from positive x-axis). $\vec{F}_2$: 15 N at $120^\circ$ (from positive x-axis). 1. **Resolve components:** $\vec{F}_1 = \langle 10 \cos 30^\circ, 10 \sin 30^\circ \rangle = \langle 8.66, 5 \rangle$ $\vec{F}_2 = \langle 15 \cos 120^\circ, 15 \sin 120^\circ \rangle = \langle -7.5, 12.99 \rangle$ 2. **Sum components for net force:** $\vec{F}_{net} = \langle 8.66 - 7.5, 5 + 12.99 \rangle = \langle 1.16, 17.99 \rangle$ 3. **Magnitude of net force:** $||\vec{F}_{net}|| = \sqrt{1.16^2 + 17.99^2} \approx 18.03 \text{ N}$ 4. **Direction of net force:** $\tan \theta = \frac{17.99}{1.16} \approx 15.5086 \implies \theta \approx 86.3^\circ$.