8 Draw a diagram for each question. 8.1 If \( \tan 28^{\circ}=p \), write down the following in terms of \( p \). 8.1.1 \( \tan 152^{\circ} \) 8.1.2 \( \cos \left(-28^{\circ}\right) \) \( 8.1 .3 \sin 62^{\circ} \) 8.2 If \( 2 \sin 41^{\circ}=a \), write down the following in terms of \( a \). \( 8.2 .1 \sin 319^{\circ} \) 8. \( 2.2 \cos 49^{\circ} \) 8.2.3 \( \cos 139^{\circ} \)

Let's solve the problem step by step, starting with the first part. ### 8.1 If \( \tan 28^{\circ} = p \) #### 8.1.1 Find \( \tan 152^{\circ} \) Using the tangent identity: \[ \tan(180^{\circ} - x) = -\tan(x) \] we can express \( \tan 152^{\circ} \) as: \[ \tan 152^{\circ} = \tan(180^{\circ} - 28^{\circ}) = -\tan 28^{\circ} = -p \] #### 8.1.2 Find \( \cos(-28^{\circ}) \) Using the cosine identity: \[ \cos(-x) = \cos(x) \] we have: \[ \cos(-28^{\circ}) = \cos(28^{\circ}) \] Using the identity \( \cos(x) = \frac{1}{\sqrt{1 + \tan^2(x)}} \): \[ \cos(28^{\circ}) = \frac{1}{\sqrt{1 + p^2}} \] #### 8.1.3 Find \( \sin 62^{\circ} \) Using the sine identity: \[ \sin(90^{\circ} - x) = \cos(x) \] we can express \( \sin 62^{\circ} \) as: \[ \sin 62^{\circ} = \cos(28^{\circ} \] Thus: \[ \sin 62^{\circ} = \frac{1}{\sqrt{1 + p^2}} \] ### 8.2 If \( 2 \sin 41^{\circ} = a \) #### 8.2.1 Find \( \sin 319^{\circ} \) Using the sine identity: \[ \sin(360^{\circ} - x) = -\sin(x) \] we can express \( \sin 319^{\circ} \) as: \[ \sin 319^{\circ} = \sin(360^{\circ} - 41^{\circ}) = -\sin 41^{\circ} \] Thus: \[ \sin 319^{\circ} = -\frac{a}{2} \] #### 8.2.2 Find \( \cos 49^{\circ} \) Using the cosine identity: \[ \cos(90^{\circ} - x) = \sin(x) \] we can express \( \cos 49^{\circ} \) as: \[ \cos 49^{\circ} = \sin(41^{\circ} \] Thus: \[ \cos 49^{\circ} = \frac{a}{2} \] #### 8.2.3 Find \( \cos 139^{\circ} \) Using the cosine identity: \[ \cos(180^{\circ} - x) = -\cos(x) \] we can express \( \cos 139^{\circ} \) as: \[ \cos 139^{\circ} = -\cos(41^{\circ}) \] Using the identity \( \cos(41^{\circ}) = \sqrt{1 - \sin^2(41^{\circ})} \): \[ \cos(41^{\circ}) = \sqrt{1 - \left(\frac{a}{2}\right)^2} \] Thus: \[ \cos 139^{\circ} = -\sqrt{1 - \left(\frac{a}{2}\right)^2} \] ### Summary of Results 1. \( \tan 152^{\circ} = -p \) 2. \( \cos(-28^{\circ}) = \frac{1}{\sqrt{1 + p^2}} \) 3. \( \sin 62^{\circ} = \frac{1}{\sqrt{1 + p^2}} \) 4. \( \sin 319^{\circ} = -\frac{a}{2} \) 5. \( \cos 49^{\circ} = \frac{a}{2} \) 6. \( \cos 139^{\circ} = -\sqrt{1 - \left(\frac{a}{2}\right)^2} \) This concludes the solution to the problem. If you need diagrams or further explanations, please let me know!