If \( 2 \sin \theta+7=0 \) and \( 90^{\circ}<\theta<270^{\circ} \), Determine the value of \( 24 \tan \theta+25 \cos \theta \) with the aid of a diagram. If \( \cos 20^{\circ}=p \), express each of the following in terms of \( p \) : \( 5.2 .1 \cos \left(-20^{\circ}\right) \) \( 5.2 .2 \cos 160^{\circ} \)
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To solve \( 2 \sin \theta + 7 = 0 \), we first isolate \( \sin \theta \) by moving \( 7 \) to the other side: \[ 2 \sin \theta = -7 \implies \sin \theta = -\frac{7}{2} \] However, this value is outside the possible range for the sine function, which is from \(-1\) to \(1\). So, let's double-check the initial interpretation. If we correctly interpret \( \theta \) as an angle, we've actually made a mistake; since \( \sin \theta \) cannot equal \(-\frac{7}{2}\), we need to review the context. If we consider \( \theta \) in the range \(90^{\circ} < \theta < 270^{\circ}\), we might have made an error in solving for the sine function, as \(2 \sin \theta + 7 = 0\) doesn’t yield a valid sine value. In essence, let’s focus on what the problem requires regarding \( 24 \tan \theta + 25 \cos \theta \): since \( \sin \theta\) is not valid, we won’t be able to compute that expression. Now, expressing terms in terms of \(p\): \[ 5.2.1 \cos(-20^{\circ}) = \cos(20^{\circ}) = p \] \[ 5.2.2 \cos(160^{\circ}) = -\cos(20^{\circ}) = -p \] So, specifically, the answers are \(5.2.1\) gives \(5.2 \cdot 1 \cdot p\) and \(5.2.2\) gives \(5.2 \cdot (-p)\) respectively.