1. State the problem. 4.2.1 Solve $\sin\theta + 0.38 = 1$ for $\theta$, correct to one decimal place. 2. Formula and rules. We isolate the sine by using $\sin\theta = 1 - 0.38$. We remember that if $\sin\theta = s$ then principal solution is $\theta = \arcsin(s)$ and in $0^\circ\le\theta<360^\circ$ the other solution is $\theta = 180^\circ - \arcsin(s)$. 3. Intermediate work. Compute $\sin\theta = 0.62$. Take the inverse sine: $\theta = \arcsin(0.62)$. Numerically $\theta \approx 38.3^\circ$ to one decimal place. The second solution in $[0^\circ,360^\circ)$ is $\theta = 180^\circ - 38.3^\circ = 141.7^\circ$. 4. Final answer for 4.2.1. $\theta \approx 38.3^\circ$ or $141.7^\circ$ (to one decimal place). 1. State the problem. 4.2.2 Solve $2\cot 2\theta = 3$ for $\theta$, correct to one decimal place. 2. Formula and rules. First isolate cotangent: $\cot 2\theta = \frac{3}{2}$. Use the reciprocal identity $\cot x = \frac{1}{\tan x}$ so $\tan 2\theta = \frac{2}{3}$. General solution for $\tan x = t$ is $x = \arctan(t) + 180^\circ k$. 3. Intermediate work. Compute $2\theta = \arctan\!\left(\frac{2}{3}\right) + 180^\circ k$. Numerically $\arctan\!\left(\frac{2}{3}\right) \approx 33.690^\circ$. Thus $2\theta \approx 33.690^\circ + 180^\circ k$. Divide by 2 to get $\theta \approx 16.845^\circ + 90^\circ k$. Rounded to one decimal place $\theta \approx 16.8^\circ + 90^\circ k$. 4. Final answer for 4.2.2. $\theta \approx 16.8^\circ + 90^\circ k$ for integer $k$, so in $[0^\circ,360^\circ)$ the values are $16.8^\circ,106.8^\circ,196.8^\circ,286.8^\circ$ (to one decimal place). 1. State the problem. 4.2.3 Solve $2\cos(3\theta - 60^\circ) = 1.71$ for $\theta$, correct to one decimal place. 2. Formula and rules. Isolate cosine: $\cos(3\theta - 60^\circ)=0.855$. For $\cos x = a$ the general solution is $x = \arccos(a) + 360^\circ k$ or $x = -\arccos(a) + 360^\circ k$. 3. Intermediate work. Compute $\alpha = \arccos(0.855) \approx 31.2^\circ$. Thus $3\theta - 60^\circ = \pm 31.2^\circ + 360^\circ k$. So $3\theta = 60^\circ \pm 31.2^\circ + 360^\circ k$. Divide by 3 to get two families: $\theta \approx \dfrac{60^\circ + 31.2^\circ}{3} = 30.4^\circ + 120^\circ k$ and $\theta \approx \dfrac{60^\circ - 31.2^\circ}{3} = 9.6^\circ + 120^\circ k$. Rounded to one decimal place the solutions are $\theta \approx 30.4^\circ + 120^\circ k$ and $9.6^\circ + 120^\circ k$. 4. Final answer for 4.2.3. $\theta \approx 9.6^\circ + 120^\circ k$ or $30.4^\circ + 120^\circ k$ for integer $k$ (values shown to one decimal place). 1. State the problem. 5.1 Given $\sin\theta = \dfrac{4}{5}$ and $\tan\theta<0$, determine $\tan\theta$ without a calculator. 2. Formula and rules. Use a right triangle with hypotenuse $5$ and opposite side $4$ so adjacent is $\sqrt{5^2-4^2}=3$ in magnitude. Sign of $\tan\theta$ is negative, and since $\sin\theta>0$ while $\tan\theta<0$, $\cos\theta$ must be negative so the adjacent side is $-3$ (quadrant II). Recall $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$. 3. Intermediate work. Compute $\cos\theta = -\dfrac{3}{5}$. Therefore $\tan\theta = \dfrac{4/5}{-3/5} = -\dfrac{4}{3}$. 4. Final answer for 5.1.1. $\tan\theta = -\dfrac{4}{3}$. 1. State the problem. 5.1.2 Compute $2\cos^{2}\theta - 1$ given the same information, without a calculator. 2. Formula and rules. Use the identity $2\cos^{2}\theta - 1 = \cos 2\theta$ and compute directly from $\cos\theta = -\dfrac{3}{5}$. 3. Intermediate work. Compute $\cos^{2}\theta = \left(-\dfrac{3}{5}\right)^{2} = \dfrac{9}{25}$. Then $2\cos^{2}\theta - 1 = 2\cdot\dfrac{9}{25} - 1 = \dfrac{18}{25} - 1 = -\dfrac{7}{25}$. 4. Final answer for 5.1.2. $2\cos^{2}\theta - 1 = -\dfrac{7}{25}$. 1. State the problem. 5.2 Simplify $\cos 0^\circ + \sin^{2}60^\circ + \sqrt{2}\cdot\sec 45^\circ$ without a calculator. 2. Formula and rules. Recall values: $\cos 0^\circ = 1$, $\sin 60^\circ = \dfrac{\sqrt{3}}{2}$ so $\sin^{2}60^\circ = \dfrac{3}{4}$, and $\sec 45^\circ = \dfrac{1}{\cos 45^\circ} = \sqrt{2}$. 3. Intermediate work. Compute $\sqrt{2}\cdot\sec 45^\circ = \sqrt{2}\cdot\sqrt{2} = 2$. Sum all terms: $1 + \dfrac{3}{4} + 2 = 3 + \dfrac{3}{4} = \dfrac{15}{4}$. 4. Final answer for 5.2. $\cos 0^\circ + \sin^{2}60^\circ + \sqrt{2}\cdot\sec 45^\circ = \dfrac{15}{4}$. 1. State the problem. 5.3 A person at A is 30 m from B horizontally, the angle of elevation to window C is $15^\circ$ and to window D is $21^\circ$. Determine the vertical distance between the two windows. 2. Formula and rules. If the horizontal distance is $30$ m and the angles of elevation are $\alpha$ and $\beta$, the respective heights above ground are $30\tan\alpha$ and $30\tan\beta$. The distance between the windows is the difference $30(\tan\beta-\tan\alpha)$. 3. Intermediate work. Let $h_C=30\tan 15^\circ$ and $h_D=30\tan 21^\circ$. So the vertical separation is $\Delta h = 30\bigl(\tan 21^\circ - \tan 15^\circ\bigr)$. Numerically $\tan 21^\circ \approx 0.3839$ and $\tan 15^\circ \approx 0.2679$. Thus $\Delta h \approx 30(0.3839-0.2679)=30(0.1160)\approx 3.48$ m. 4. Final answer for 5.3. The distance between the two windows is $30\bigl(\tan 21^\circ-\tan 15^\circ\bigr)\approx 3.48$ m.