1. **Problem 1: Calculate the exact length of AB in △ABC and △ACD** Given: - △ABC is right-angled with $\angle BAC = 45^\circ$ and $\angle BCA = 45^\circ$. - △ACD is constructed on hypotenuse AD with $\angle ACD = 90^\circ$, $CD = 5$ cm, and $\angle ADC = 60^\circ$. Since $\angle BAC = \angle BCA = 45^\circ$, triangle ABC is isosceles right triangle, so $AB = BC$. 2. **Find length AD (hypotenuse of △ABC):** In △ACD, $\angle ACD = 90^\circ$, $\angle ADC = 60^\circ$, so $\angle CAD = 30^\circ$ (sum of angles in triangle = 180°). Using right triangle ACD with $CD = 5$ cm opposite $30^\circ$, hypotenuse AD is: $$AD = \frac{CD}{\sin 30^\circ} = \frac{5}{\frac{1}{2}} = 10\text{ cm}$$ 3. **Find length AC:** Using $\cos 30^\circ = \frac{AC}{AD}$, $$AC = AD \times \cos 30^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}\text{ cm}$$ 4. **Find length AB:** Since △ABC is isosceles right triangle with hypotenuse AD = 10 cm, $$AB = BC = \frac{AD}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2}\text{ cm}$$ --- 1. **Problem 2: Find hypotenuse given opposite side 13 cm and $\tan \theta = 1$** Given: - Opposite side = 13 cm - $\tan \theta = 1$ means opposite = adjacent 2. **Find adjacent side:** Since $\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = 1$, adjacent side = 13 cm. 3. **Find hypotenuse using Pythagoras:** $$\text{hypotenuse} = \sqrt{13^2 + 13^2} = \sqrt{169 + 169} = \sqrt{338} = 13\sqrt{2}$$ --- 1. **Problem 3: Calculate exact distance between sniper and target** Given: - Horizontal distance between target and point A = 50 m - Angle of depression from sniper to target = 45° - Angle of elevation from target to sniper = 30° - Distance between point A and B equals sniper to B vertically 2. **Find height AB:** From angle of depression 45°, height AB = horizontal distance = 50 m (since tan 45° = 1). 3. **Find distance from sniper to B:** Since AB = sniper to B vertically, AB = 50 m. 4. **Find distance from sniper to target:** Using angle of elevation 30° at target, $$\tan 30^\circ = \frac{AB}{\text{distance sniper to target}} \Rightarrow \text{distance} = \frac{AB}{\tan 30^\circ} = \frac{50}{\frac{1}{\sqrt{3}}} = 50\sqrt{3}$$ --- 1. **Problem 4: Simplify expression** $$\tan^2 60^\circ \times \frac{\sin^2 30^\circ - \cos^2 60^\circ}{\cos^2 30^\circ + \sin^2 60^\circ}$$ 2. **Calculate each term:** - $\tan 60^\circ = \sqrt{3} \Rightarrow \tan^2 60^\circ = 3$ - $\sin 30^\circ = \frac{1}{2} \Rightarrow \sin^2 30^\circ = \frac{1}{4}$ - $\cos 60^\circ = \frac{1}{2} \Rightarrow \cos^2 60^\circ = \frac{1}{4}$ - $\cos 30^\circ = \frac{\sqrt{3}}{2} \Rightarrow \cos^2 30^\circ = \frac{3}{4}$ - $\sin 60^\circ = \frac{\sqrt{3}}{2} \Rightarrow \sin^2 60^\circ = \frac{3}{4}$ 3. **Simplify numerator:** $$\sin^2 30^\circ - \cos^2 60^\circ = \frac{1}{4} - \frac{1}{4} = 0$$ 4. **Simplify denominator:** $$\cos^2 30^\circ + \sin^2 60^\circ = \frac{3}{4} + \frac{3}{4} = \frac{6}{4} = \frac{3}{2}$$ 5. **Final value:** $$3 \times \frac{0}{\frac{3}{2}} = 0$$ --- 1. **Problem 5: Given $\sin \theta = -\frac{5}{13}$, $270^\circ < \theta < 360^\circ$, find $\cos \theta$ and $\tan \theta$** 2. **Find $\cos \theta$:** Using Pythagorean identity: $$\cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(-\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169}$$ Since $270^\circ < \theta < 360^\circ$ (4th quadrant), $\cos \theta > 0$, so: $$\cos \theta = \frac{12}{13}$$ 3. **Find $\tan \theta$:** $$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{5}{13}}{\frac{12}{13}} = -\frac{5}{12}$$ --- 1. **Problem 6: Evaluate $\cos(-660^\circ)$** 2. **Reduce angle:** $$-660^\circ + 720^\circ = 60^\circ$$ So, $$\cos(-660^\circ) = \cos 60^\circ = \frac{1}{2}$$ --- 1. **Problem 7: Simplify $\cos(\pi + \theta) - \sin\left(\frac{\pi}{2} + \theta\right)$ in terms of $\cos \theta$** 2. **Use identities:** - $\cos(\pi + \theta) = -\cos \theta$ - $\sin\left(\frac{\pi}{2} + \theta\right) = \cos \theta$ 3. **Simplify:** $$-\cos \theta - \cos \theta = -2 \cos \theta$$ --- **Final answers:** - Length AB = $5\sqrt{2}$ cm - Hypotenuse = $13\sqrt{2}$ cm - Distance sniper to target = $50\sqrt{3}$ m - Expression value = 0 - $\cos \theta = \frac{12}{13}$, $\tan \theta = -\frac{5}{12}$ - $\cos(-660^\circ) = \frac{1}{2}$ - Simplified expression = $-2 \cos \theta$