2. Without using a calculator, state whether the following ratios are positive or negative: $$ \begin{array}{lll} ext { a) } \cos 325^{\circ} & ext { b) } an 126^{\circ} & ext { c) } \sin 154^{\circ}\end{array} $$ 3. In which quadrant will $$ x $$ lie if: $$ \begin{array}{ll} ext { a) } \sin x0 ext { and } an x0 ext { and } \cos x

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2. Without using a calculator, state whether the following ratios are positive or negative: $$ \begin{array}{lll} ext { a) } \cos 325^{\circ} & ext { b) } an 126^{\circ} & ext { c) } \sin 154^{\circ}\end{array} $$ 3. In which quadrant will $$ x $$ lie if: $$ \begin{array}{ll} ext { a) } \sin x>0 ext { and } an x<0 & ext { b) } an x>0 ext { and } \cos x<0 \ ext { c) } an x<0 ext { and } 180^{\circ}0 ext { and } \sin x<0 ext { ? }\end{array} $$ 4. If $$ 12 an C+5=0 $$ and $$ C \in\left(90^{\circ} ; 270^{\circ} ight) $$, find the value of $$ 12 an C-13 \cos C $$ with the aid of a diagram and without using a calculator.

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Alright, I need to tackle these trigonometry problems without using a calculator. Let's break them down one by one. **Problem 2:** a) $$ \cos 325^{\circ} $$ b) $$ an 126^{\circ} $$ c) $$ \sin 154^{\circ} $$ First, I recall that cosine and sine functions have specific signs in each quadrant: - **First Quadrant (0°-90°):** All trigonometric functions are positive. - **Second Quadrant (90°-180°):** Sine is positive, cosine and tangent are negative. - **Third Quadrant (180°-270°):** Tangent is positive, sine and cosine are negative. - **Fourth Quadrant (270°-360°):** Cosine is positive, sine and tangent are negative. For **a) $$ \cos 325^{\circ} $$:** 325° is in the fourth quadrant. In the fourth quadrant, cosine is positive. So, $$ \cos 325^{\circ} $$ is **positive**. For **b) $$ an 126^{\circ} $$:** 126° is in the second quadrant. In the second quadrant, tangent is negative. So, $$ an 126^{\circ} $$ is **negative**. For **c) $$ \sin 154^{\circ} $$:** 154° is in the second quadrant. In the second quadrant, sine is positive. So, $$ \sin 154^{\circ} $$ is **positive**. **Problem 3:** a) $$ \sin x > 0 $$ and $$ an x < 0 $$ b) $$ an x > 0 $$ and $$ \cos x < 0 $$ c) $$ an x < 0 $$ and $$ 180^{\circ} < x < 360^{\circ} $$ d) $$ \cos x > 0 $$ and $$ \sin x < 0 $$ Let's analyze each case: a) $$ \sin x > 0 $$ and $$ an x < 0 $$: - $$ \sin x > 0 $$ in the first and second quadrants. - $$ an x < 0 $$ in the second and fourth quadrants. - The only quadrant where both conditions are satisfied is the **second quadrant**. b) $$ an x > 0 $$ and $$ \cos x < 0 $$: - $$ an x > 0 $$ in the first and third quadrants. - $$ \cos x < 0 $$ in the second and third quadrants. - The only quadrant where both conditions are satisfied is the **third quadrant**. c) $$ an x < 0 $$ and $$ 180^{\circ} < x < 360^{\circ} $$: - $$ an x < 0 $$ in the second and fourth quadrants. - The range $$ 180^{\circ} < x < 360^{\circ} $$ covers the third and fourth quadrants. - $$ an x < 0 $$ in the **fourth quadrant**. d) $$ \cos x > 0 $$ and $$ \sin x < 0 $$: - $$ \cos x > 0 $$ in the first and fourth quadrants. - $$ \sin x < 0 $$ in the third and fourth quadrants. - The only quadrant where both conditions are satisfied is the **fourth quadrant**. **Problem 4:** Given $$ 12 an C + 5 = 0 $$ and $$ C \in (90^{\circ}, 270^{\circ}) $$, find the value of $$ 12 an C - 13 \cos C $$ with the aid of a diagram and without using a calculator. First, solve for $$ an C $$: $$ 12 an C + 5 = 0 $$ $$ 12 an C = -5 $$ $$ an C = -\frac{5}{12} $$ Since $$ C $$ is in the second quadrant (90° < C < 270°), and tangent is negative there, this makes sense. Now, find $$ \cos C $$. Using the identity: $$ an C = \frac{\sin C}{\cos C} $$ $$ -\frac{5}{12} = \frac{\sin C}{\cos C} $$ $$ \sin C = -\frac{5}{12} \cos C $$ Using the Pythagorean identity: $$ \sin^2 C + \cos^2 C = 1 $$ $$ \left(-\frac{5}{12} \cos C ight)^2 + \cos^2 C = 1 $$ $$ \frac{25}{144} \cos^2 C + \cos^2 C = 1 $$ $$ \frac{169}{144} \cos^2 C = 1 $$ $$ \cos^2 C = \frac{144}{169} $$ $$ \cos C = \pm \frac{12}{13} $$ Since $$ C $$ is in the second quadrant, cosine is negative: $$ \cos C = -\frac{12}{13} $$ Now, find $$ \sin C $$: $$ \sin C = -\frac{5}{12} \cos C = -\frac{5}{12} imes -\frac{12}{13} = \frac{5}{13} $$ Now, compute $$ 12 an C - 13 \cos C $$: $$ 12 an C = 12 imes -\frac{5}{12} = -5 $$ $$ -13 \cos C = -13 imes -\frac{12}{13} = 12 $$ So, $$ -5 + 12 = 7 $$ Therefore, the value is **7**. 2. a) $$ \cos 325^{\circ} $$ is **positive**. b) $$ an 126^{\circ} $$ is **negative**. c) $$ \sin 154^{\circ} $$ is **positive**. 3. a) $$ x $$ lies in the **second quadrant**. b) $$ x $$ lies in the **third quadrant**. c) $$ x $$ lies in the **fourth quadrant**. d) $$ x $$ lies in the **fourth quadrant**. 4. The value of $$ 12 an C - 13 \cos C $$ is **7**.

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