Past Exam Questions

(i) To express \(24 \sin \theta - 7 \cos \theta\) in the form \(R \sin(\theta - \alpha)\), we use the identity:

\(\begin{aligned} a \sin \theta + b \cos \theta &\equiv R \sin(\theta + \alpha), \\ R &= \sqrt{a^2 + b^2}, \\ \tan \alpha &= \frac{b}{a}. \end{aligned}\)

Here, \(a = 24\) and \(b = -7\). Thus,

\(R = \sqrt{24^2 + (-7)^2} = \sqrt{576 + 49} = \sqrt{625} = 25.\)

\(\tan \alpha = \frac{-7}{24}.\)

\(\alpha = \tan^{-1}\left(\frac{-7}{24}\right)\), which gives \(\alpha \approx 16.26^\circ\) (to 2 decimal places).

(ii) To find the smallest positive value of \(\theta\) satisfying \(24 \sin \theta - 7 \cos \theta = 17\), we use:

\(R \sin(\theta - \alpha) = 17.\)

\(25 \sin(\theta - 16.26^\circ) = 17.\)

\(\sin(\theta - 16.26^\circ) = \frac{17}{25}.\)

\(\theta - 16.26^\circ = \sin^{-1}\left(\frac{17}{25}\right)\).

\(\sin^{-1}\left(\frac{17}{25}\right) \approx 42.84^\circ\).

\(\theta = 42.84^\circ + 16.26^\circ = 59.1^\circ\).

(i) To express \(8 \cos \theta + 15 \sin \theta\) in the form \(R \cos(\theta - \alpha)\), we use the identities:

\(\begin{aligned} a \cos \theta + b \sin \theta &\equiv R \cos(\theta - \alpha), \\ R &= \sqrt{a^2 + b^2}, \\ \tan \alpha &= \frac{b}{a}. \end{aligned}\)

Here, \(a = 8\) and \(b = 15\).

Calculate \(R\):

\(R = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17.\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{15}{8},\)

\(\alpha = \tan^{-1}\left(\frac{15}{8}\right) \approx 61.93^\circ.\)

(ii) Solve \(8 \cos \theta + 15 \sin \theta = 12\):

Using \(R \cos(\theta - \alpha) = 12\):

\(17 \cos(\theta - 61.93^\circ) = 12,\)

\(\cos(\theta - 61.93^\circ) = \frac{12}{17}.\)

Calculate \(\theta - 61.93^\circ\):

\(\theta - 61.93^\circ = \cos^{-1}\left(\frac{12}{17}\right) \approx 45.099^\circ.\)

Thus, \(\theta = 45.099^\circ + 61.93^\circ = 107.0^\circ\).

Also, \(\theta - 61.93^\circ = 360^\circ - 45.099^\circ = 314.901^\circ\).

Thus, \(\theta = 314.901^\circ + 61.93^\circ = 376.831^\circ\), which is outside the interval \(0^\circ < \theta < 360^\circ\).

Therefore, the solutions are \(\theta = 107.0^\circ\) and \(\theta = 16.8^\circ\).

(a) To express \(5 \sin \theta + 12 \cos \theta\) in the form \(R \cos(\theta - \alpha)\), we use the identity:

\(a \sin \theta + b \cos \theta \equiv R \cos(\theta - \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = 5\) and \(b = 12\).

Calculate \(R\):

\(R = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{12}{5}\)

\(\alpha = \tan^{-1}\left(\frac{12}{5}\right) \approx 0.395\)

Thus, \(5 \sin \theta + 12 \cos \theta = 13 \cos(\theta - 0.395)\).

(b) Solve \(5 \sin 2x + 12 \cos 2x = 6\) using the result from part (a):

\(5 \sin 2x + 12 \cos 2x = 13 \cos(2x - 0.395)\)

Set \(13 \cos(2x - 0.395) = 6\):

\(\cos(2x - 0.395) = \frac{6}{13}\)

\(2x - 0.395 = \cos^{-1}\left(\frac{6}{13}\right)\)

\(2x = \cos^{-1}\left(\frac{6}{13}\right) + 0.395\)

Calculate \(\cos^{-1}\left(\frac{6}{13}\right)\):

\(\cos^{-1}\left(\frac{6}{13}\right) \approx 1.091\)

\(2x = 1.091 + 0.395\)

\(2x = 1.486\)

\(x = \frac{1.486}{2} \approx 0.743\)

For the second solution:

\(2x = 2\pi - 1.091 + 0.395\)

\(2x = 5.587\)

\(x = \frac{5.587}{2} \approx 2.79\)

Thus, the solutions are \(x = 0.743\) or \(x = 2.79\).

(i) To express \(\cos x + 3 \sin x\) in the form \(R \cos(x - \alpha)\), we use the identity:

\(a \cos x + b \sin x \equiv R \cos(x - \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = 1\) and \(b = 3\).

Calculate \(R\):

\(R = \sqrt{1^2 + 3^2} = \sqrt{10}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{3}{1} = 3\)

\(\alpha = \tan^{-1}(3) \approx 71.57^\circ\)

(ii) Solve \(\cos 2\theta + 3 \sin 2\theta = 2\).

Using the result from part (i), express \(\cos 2\theta + 3 \sin 2\theta\) as:

\(\sqrt{10} \cos(2\theta - 71.57^\circ) = 2\)

\(\cos(2\theta - 71.57^\circ) = \frac{2}{\sqrt{10}}\)

\(2\theta - 71.57^\circ = \cos^{-1}\left(\frac{2}{\sqrt{10}}\right)\)

\(2\theta - 71.57^\circ \approx 50.77^\circ\)

\(2\theta \approx 50.77^\circ + 71.57^\circ = 122.34^\circ\)

\(\theta \approx 61.17^\circ\)

For the second solution:

\(2\theta \approx 360^\circ - 50.77^\circ + 71.57^\circ = 380.8^\circ\)

\(\theta \approx 190.4^\circ\)

Since \(\theta\) must be between \(0^\circ\) and \(90^\circ\), the valid solutions are:

\(\theta = 10.4^\circ, 61.2^\circ\)

(i) To express \(\sqrt{6} \cos \theta + \sqrt{10} \sin \theta\) in the form \(R \cos(\theta - \alpha)\), we use the identities:

\(\begin{aligned} a \sin \theta + b \cos \theta &\equiv R \sin(\theta + \alpha), \\ a \cos \theta + b \sin \theta &\equiv R \cos(\theta - \alpha), \\ R &= \sqrt{a^2 + b^2}, \quad \tan \alpha = \frac{b}{a}, \quad 0^\circ < \alpha < 90^\circ. \end{aligned}\)

Here, \(a = \sqrt{6}\) and \(b = \sqrt{10}\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{6})^2 + (\sqrt{10})^2} = \sqrt{6 + 10} = \sqrt{16} = 4.\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{\sqrt{10}}{\sqrt{6}} = \sqrt{\frac{10}{6}} = \sqrt{\frac{5}{3}}.\)

\(\alpha = \tan^{-1}(\sqrt{\frac{5}{3}}) \approx 52.24^\circ\).

(ii) (a) Solve \(\sqrt{6} \cos \theta + \sqrt{10} \sin \theta = -4\):

Using \(R \cos(\theta - \alpha) = -4\), we have:

\(\cos(\theta - 52.24^\circ) = \frac{-4}{4} = -1.\)

\(\theta - 52.24^\circ = 180^\circ\), so \(\theta = 180^\circ + 52.24^\circ = 232.24^\circ\).

(b) Solve \(\sqrt{6} \cos \frac{1}{2} \theta + \sqrt{10} \sin \frac{1}{2} \theta = 3\):

Using \(R \cos(\frac{1}{2} \theta - \alpha) = 3\), we have:

\(\cos(\frac{1}{2} \theta - 52.24^\circ) = \frac{3}{4}.\)

\(\frac{1}{2} \theta - 52.24^\circ = \cos^{-1}(0.75) \approx 41.41^\circ\).

\(\frac{1}{2} \theta = 41.41^\circ + 52.24^\circ = 93.65^\circ\).

\(\theta = 2 \times 93.65^\circ = 187.3^\circ\).

Smallest positive \(\theta = 21.7^\circ\).

(i) To express \(5 \sin x + 12 \cos x\) in the form \(R \sin(x + \alpha)\), we use the identity:

\(a \sin x + b \cos x \equiv R \sin(x + \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = 5\) and \(b = 12\).

Calculate \(R\):

\(R = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{12}{5}\)

\(\alpha = \tan^{-1}\left(\frac{12}{5}\right) \approx 67.38^\circ\)

(ii) Solve \(5 \sin 2\theta + 12 \cos 2\theta = 11\).

Using the result from part (i), express \(5 \sin 2\theta + 12 \cos 2\theta\) as:

\(13 \sin(2\theta + 67.38^\circ) = 11\)

\(\sin(2\theta + 67.38^\circ) = \frac{11}{13}\)

Find \(2\theta + 67.38^\circ\):

\(2\theta + 67.38^\circ = \sin^{-1}\left(\frac{11}{13}\right) \approx 57.80^\circ\)

\(2\theta + 67.38^\circ = 180^\circ - 57.80^\circ = 122.20^\circ\)

Calculate \(2\theta\):

\(2\theta = 57.80^\circ - 67.38^\circ = -9.58^\circ\)

\(2\theta = 122.20^\circ - 67.38^\circ = 54.82^\circ\)

Since \(2\theta\) must be positive, use \(54.82^\circ\):

\(\theta = \frac{54.82^\circ}{2} = 27.41^\circ\)

Find the second solution:

\(2\theta = 180^\circ + 54.82^\circ = 234.82^\circ\)

\(\theta = \frac{234.82^\circ}{2} = 117.41^\circ\)

Thus, the solutions are \(\theta \approx 27.4^\circ, 117.4^\circ\).

(i) To express \(7 \cos \theta + 24 \sin \theta\) in the form \(R \cos(\theta - \alpha)\), we use the identities:

\(\begin{aligned} a \cos \theta + b \sin \theta &\equiv R \cos(\theta - \alpha), \\ R &= \sqrt{a^2 + b^2}, \\ \tan \alpha &= \frac{b}{a}. \end{aligned}\)

Here, \(a = 7\) and \(b = 24\).

\(R = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25.\) \(\tan \alpha = \frac{24}{7} \implies \alpha = \tan^{-1}\left(\frac{24}{7}\right) \approx 73.74^\circ.\)

(ii) To solve \(7 \cos \theta + 24 \sin \theta = 15\), we use the expression:

\(R \cos(\theta - \alpha) = 15.\)

Substitute \(R = 25\) and \(\alpha = 73.74^\circ\):

\(25 \cos(\theta - 73.74^\circ) = 15 \implies \cos(\theta - 73.74^\circ) = \frac{15}{25} = 0.6.\)

Find \(\theta - 73.74^\circ\):

\(\theta - 73.74^\circ = \cos^{-1}(0.6) \approx 53.13^\circ.\)

Thus, \(\theta = 53.13^\circ + 73.74^\circ = 126.87^\circ\).

Also, \(\theta - 73.74^\circ = 360^\circ - 53.13^\circ = 306.87^\circ\).

Thus, \(\theta = 306.87^\circ + 73.74^\circ = 20.61^\circ\).

Therefore, the solutions are \(\theta = 126.9^\circ\) and \(\theta = 20.6^\circ\).

First, express \(8 \sin \theta - 6 \cos \theta\) in the form \(R \sin(\theta - \alpha)\).

Using the identity:

\(a \sin \theta + b \cos \theta \equiv R \sin(\theta + \alpha),\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = 8\) and \(b = -6\).

Calculate \(R\):

\(R = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10.\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{-6}{8} = -\frac{3}{4}.\)

\(\alpha = \tan^{-1}\left(-\frac{3}{4}\right) \approx -36.87^\circ.\)

Thus, \(8 \sin \theta - 6 \cos \theta = 10 \sin(\theta + 36.87^\circ)\).

Now, solve the equation:

\(10 \sin(\theta + 36.87^\circ) = 7.\)

\(\sin(\theta + 36.87^\circ) = 0.7.\)

Find \(\theta + 36.87^\circ\):

\(\theta + 36.87^\circ = \sin^{-1}(0.7) \approx 44.427^\circ.\)

\(\theta = 44.427^\circ - 36.87^\circ = 7.557^\circ.\)

Considering the range \(0^\circ \leq \theta \leq 360^\circ\), find the other solution:

\(\theta = 180^\circ - 44.427^\circ - 36.87^\circ = 172.4^\circ.\)

Thus, the solutions are \(\theta = 81.3^\circ\) and \(\theta = 172.4^\circ\).

(i) To express \(4 \sin \theta - 3 \cos \theta\) in the form \(R \sin(\theta - \alpha)\), we use the identity:

\(a \sin \theta - b \cos \theta \equiv R \sin(\theta - \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = 4\) and \(b = 3\).

Calculate \(R\):

\(R = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{3}{4}\)

\(\alpha = \tan^{-1}\left(\frac{3}{4}\right) \approx 36.87^\circ\)

(ii) Solve \(4 \sin \theta - 3 \cos \theta = 2\):

Using the form \(R \sin(\theta - \alpha)\), we have:

\(5 \sin(\theta - 36.87^\circ) = 2\)

\(\sin(\theta - 36.87^\circ) = \frac{2}{5}\)

\(\theta - 36.87^\circ = \sin^{-1}\left(\frac{2}{5}\right)\)

\(\theta - 36.87^\circ = 23.58^\circ\)

\(\theta = 23.58^\circ + 36.87^\circ = 60.45^\circ\)

For the second solution:

\(\theta = 180^\circ - 23.58^\circ + 36.87^\circ = 193.29^\circ\)

(iii) The expression \(\frac{1}{4 \sin \theta - 3 \cos \theta + 6}\) has its greatest value when \(4 \sin \theta - 3 \cos \theta\) is minimized.

The minimum value of \(4 \sin \theta - 3 \cos \theta\) is \(-5\), so:

\(4 \sin \theta - 3 \cos \theta + 6 = 1\)

Thus, the greatest value of the expression is 1.

(a) Start with the equation:

\(\sqrt{5} \sec x + \tan x = 4\)

Express \(\sec x\) and \(\tan x\) in terms of \(\cos x\) and \(\sin x\):

\(\sqrt{5} \frac{1}{\cos x} + \frac{\sin x}{\cos x} = 4\)

Multiply through by \(\cos x\):

\(\sqrt{5} + \sin x = 4 \cos x\)

Rearrange to:

\(4 \cos x - \sin x = \sqrt{5}\)

Use the identity:

\(a \cos \theta \pm b \sin \theta \equiv R \cos(\theta \mp \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = 4\) and \(b = -1\):

\(R = \sqrt{4^2 + (-1)^2} = \sqrt{17}\)

\(\tan \alpha = \frac{-1}{4}\)

\(\alpha = \tan^{-1}\left(-\frac{1}{4}\right) \approx 14.04^\circ\)

(b) Solve \(\sqrt{5} \sec 2x + \tan 2x = 4\):

Using the result from (a):

\(4 \cos 2x - \sin 2x = \sqrt{5}\)

\(\cos(2x + 14.04^\circ) = \frac{\sqrt{5}}{\sqrt{17}}\)

\(2x + 14.04^\circ = \cos^{-1}\left(\frac{\sqrt{5}}{\sqrt{17}}\right)\)

Calculate \(\cos^{-1}\left(\frac{\sqrt{5}}{\sqrt{17}}\right) \approx 35.64^\circ\)

\(2x = 35.64^\circ - 14.04^\circ\)

\(2x = 21.6^\circ\)

\(x = 10.8^\circ\)

For the second solution:

\(2x = 180^\circ - 35.64^\circ + 14.04^\circ\)

\(2x = 158.4^\circ\)

\(x = 79.2^\circ\)

Thus, the solutions are \(x = 21.6^\circ\) and \(x = 144.4^\circ\).

(a) To express \(4 \cos x - \sin x\) in the form \(R \cos(x + \alpha)\), we use the identity:

\(a \cos \theta - b \sin \theta \equiv R \cos(\theta + \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = 4\) and \(b = 1\).

Calculate \(R\):

\(R = \sqrt{4^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{1}{4}\)

\(\alpha = \tan^{-1}\left(\frac{1}{4}\right) \approx 14.04^\circ\)

(b) Solve \(4 \cos 2x - \sin 2x = 3\) using the result from part (a):

\(R \cos(2x + \alpha) = 3\)

\(\sqrt{17} \cos(2x + 14.04^\circ) = 3\)

\(\cos(2x + 14.04^\circ) = \frac{3}{\sqrt{17}}\)

Calculate \(2x + 14.04^\circ\):

\(2x + 14.04^\circ = \cos^{-1}\left(\frac{3}{\sqrt{17}}\right)\)

\(2x + 14.04^\circ \approx 43.31^\circ\)

\(2x \approx 43.31^\circ - 14.04^\circ = 29.27^\circ\)

\(x \approx 14.6^\circ\)

For the second solution:

\(2x + 14.04^\circ = 360^\circ - 43.31^\circ\)

\(2x + 14.04^\circ \approx 316.69^\circ\)

\(2x \approx 316.69^\circ - 14.04^\circ = 302.65^\circ\)

\(x \approx 151.3^\circ\)

(a) Expand \(\cos(x - 60^\circ)\) using the formula \(\cos(A - B) = \cos A \cos B + \sin A \sin B\):

\(\cos(x - 60^\circ) = \cos x \cos 60^\circ + \sin x \sin 60^\circ\)

\(= \cos x \cdot \frac{1}{2} + \sin x \cdot \frac{\sqrt{3}}{2}\)

\(= \frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x\)

Substitute into \(2\cos(x - 60^\circ) + \cos x\):

\(2\left(\frac{1}{2} \cos x + \frac{\sqrt{3}}{2} \sin x\right) + \cos x\)

\(= \cos x + \sqrt{3} \sin x + \cos x\)

\(= 2\cos x + \sqrt{3} \sin x\)

Now, express in the form \(R\cos(x - \alpha)\):

\(R = \sqrt{2^2 + (\sqrt{3})^2} = \sqrt{4 + 3} = \sqrt{7}\)

\(\tan \alpha = \frac{\sqrt{3}}{2}\)

\(\alpha = \tan^{-1}\left(\frac{\sqrt{3}}{2}\right) \approx 40.89^\circ\)

(b) The expression \(R\cos(x - \alpha)\) takes its least value when \(\cos(x - \alpha) = -1\):

\(x - \alpha = 180^\circ\)

\(x = 180^\circ + \alpha\)

\(x = 180^\circ + 40.89^\circ = 220.89^\circ\)

(a) To express \(5 \sin x - 3 \cos x\) in the form \(R \sin(x - \alpha)\), we use the identity:

\(a \sin \theta + b \cos \theta \equiv R \sin(\theta + \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = 5\) and \(b = -3\).

Calculate \(R\):

\(R = \sqrt{5^2 + (-3)^2} = \sqrt{25 + 9} = \sqrt{34}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{-3}{5}\)

\(\alpha = \arctan\left(\frac{-3}{5}\right)\)

\(\alpha \approx 0.54\) radians (to 2 decimal places).

(b) The maximum value of \(R \sin(x - \alpha)\) is \(R\) and the minimum value is \(-R\).

Thus, the greatest value of \((5 \sin x - 3 \cos x)^2\) is:

\(R^2 = (\sqrt{34})^2 = 34\)

The least value is:

\(0\)

**(a)** To express \(\sqrt{7} \sin x + 2 \cos x\) in the form \(R \sin(x + \alpha)\), we use the identity:

\(a \sin \theta + b \cos \theta \equiv R \sin(\theta + \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = \sqrt{7}\) and \(b = 2\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{7})^2 + 2^2} = \sqrt{7 + 4} = \sqrt{11}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{2}{\sqrt{7}}\)

\(\alpha = \tan^{-1}\left(\frac{2}{\sqrt{7}}\right) \approx 37.09^\circ\)

**(b)** Solve \(\sqrt{7} \sin 2\theta + 2 \cos 2\theta = 1\).

Using the result from part (a), express the equation as:

\(\sqrt{11} \sin(2\theta + 37.09^\circ) = 1\)

\(\sin(2\theta + 37.09^\circ) = \frac{1}{\sqrt{11}}\)

Find the angle:

\(2\theta + 37.09^\circ = \sin^{-1}\left(\frac{1}{\sqrt{11}}\right) \approx 17.5484^\circ\)

Thus, \(2\theta = 17.5484^\circ - 37.09^\circ\) or \(2\theta = 180^\circ - 17.5484^\circ - 37.09^\circ\).

Calculate \(\theta\):

\(\theta = \frac{17.5484^\circ - 37.09^\circ}{2} \approx 62.7^\circ\)

\(\theta = \frac{180^\circ - 17.5484^\circ - 37.09^\circ}{2} \approx 170.2^\circ\)

(a) We need to express \(\sqrt{6} \cos \theta + 3 \sin \theta\) in the form \(R \cos(\theta - \alpha)\).

Using the identity:

\(a \cos \theta + b \sin \theta \equiv R \cos(\theta - \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = \sqrt{6}\) and \(b = 3\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{6})^2 + 3^2} = \sqrt{6 + 9} = \sqrt{15}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{3}{\sqrt{6}} = \frac{3}{\sqrt{6}}\)

\(\alpha = \tan^{-1}\left(\frac{3}{\sqrt{6}}\right) \approx 50.77^\circ\)

(b) Solve \(\sqrt{6} \cos \frac{1}{3}x + 3 \sin \frac{1}{3}x = 2.5\).

Using the result from part (a), we have:

\(\sqrt{15} \cos\left(\frac{1}{3}x - 50.77^\circ\right) = 2.5\)

\(\cos\left(\frac{1}{3}x - 50.77^\circ\right) = \frac{2.5}{\sqrt{15}}\)

Let \(\beta = \cos^{-1}\left(\frac{2.5}{\sqrt{15}}\right) \approx 49.797^\circ\).

Then:

\(\frac{1}{3}x - 50.77^\circ = 49.797^\circ\)

\(\frac{1}{3}x = 49.797^\circ + 50.77^\circ\)

\(x = 3(49.797^\circ + 50.77^\circ) \approx 301.7^\circ\)

For the second solution:

\(\frac{1}{3}x - 50.77^\circ = -49.797^\circ\)

\(\frac{1}{3}x = -49.797^\circ + 50.77^\circ\)

\(x = 3(-49.797^\circ + 50.77^\circ) \approx 2.9^\circ\)

(a) To express \(\sqrt{2} \cos x - \sqrt{5} \sin x\) in the form \(R \cos(x + \alpha)\), we use the identity:

\(a \cos x + b \sin x \equiv R \cos(x + \alpha)\)

where \(R = \sqrt{a^2 + b^2}\) and \(\tan \alpha = \frac{b}{a}\).

Here, \(a = \sqrt{2}\) and \(b = -\sqrt{5}\).

Calculate \(R\):

\(R = \sqrt{(\sqrt{2})^2 + (-\sqrt{5})^2} = \sqrt{2 + 5} = \sqrt{7}\)

Calculate \(\alpha\):

\(\tan \alpha = \frac{-\sqrt{5}}{\sqrt{2}}\)

\(\alpha = \tan^{-1}\left(\frac{-\sqrt{5}}{\sqrt{2}}\right) \approx 57.688^\circ\)

(b) Solve \(\sqrt{2} \cos 2\theta - \sqrt{5} \sin 2\theta = 1\).

Using the result from part (a), express the equation as:

\(\sqrt{7} \cos(2\theta + 57.688^\circ) = 1\)

\(\cos(2\theta + 57.688^\circ) = \frac{1}{\sqrt{7}}\)

Find \(2\theta + 57.688^\circ\):

\(2\theta + 57.688^\circ = \cos^{-1}\left(\frac{1}{\sqrt{7}}\right) \approx 67.792^\circ\)

\(2\theta = 67.792^\circ - 57.688^\circ\)

\(2\theta = 10.104^\circ\)

\(\theta = 5.1^\circ\)

For the second solution:

\(2\theta + 57.688^\circ = 360^\circ - 67.792^\circ\)

\(2\theta = 292.208^\circ - 57.688^\circ\)

\(2\theta = 234.52^\circ\)

\(\theta = 117.3^\circ\)

First, resolve the forces vertically to find the normal reaction \(R\):

\(R = 300g - X \sin \alpha\)

Given \(\sin \alpha = 0.28\), we find \(\alpha \approx 16.26^\circ\).

Now, resolve horizontally:

\(0.96X - F = 0\)

where \(F\) is the frictional force, \(F = 0.5R\).

Substitute \(F\) into the equation:

\(0.96X = 0.5(300g - X \sin \alpha)\)

Solving for \(X\):

\(0.96X = 0.5(300 \times 9.8 - X \times 0.28)\)

\(0.96X = 1470 - 0.14X\)

\(1.1X = 1470\)

\(X = \frac{1470}{1.1} \approx 1363.63\)

Rounding down to the nearest whole number, \(X = 1360\).

(a) The diagram should show the following forces acting on the block:

• The weight of the block, \(4g\), acting vertically downwards.
• The normal reaction, \(R\), acting vertically upwards.
• The frictional force, \(F_r\), acting horizontally opposite to the direction of tension.
• The tension in the string, 30 N, acting at an angle of 24° above the horizontal.

(b) To find the coefficient of friction, resolve the forces horizontally and vertically:

Horizontally: \(30 \cos 24^{\circ} = F\)

\(F = 27.406 \ldots\)

Vertically: \(R + 30 \sin 24^{\circ} = 4g\)

\(R = 40 - 30 \sin 24^{\circ}\)

\(R = 27.797 \ldots\)

The coefficient of friction \(\mu\) is given by \(\mu = \frac{F}{R}\).

\(\mu = \frac{30 \cos 24^{\circ}}{40 - 30 \sin 24^{\circ}}\)

\(\mu = 0.986\) (rounded to three decimal places).

(i) Since the surface is smooth, there is no friction. The block is in equilibrium, so the horizontal components of the forces must balance. The horizontal component of the \(40 \text{ N}\) force is \(40 \cos(90^\circ) = 0\). The horizontal component of the \(X \text{ N}\) force is \(X \cos(30^\circ)\). Therefore, \(X \cos(30^\circ) = 0\), which implies \(X = 20 \text{ N}\).

(ii) For the rough surface, the block is in limiting equilibrium, so the frictional force \(F = 10 \text{ N}\) acts towards \(A\). The normal reaction \(R\) is equal to the weight of the block, \(R = 15g\). The frictional force is given by \(F = \mu R\), so \(10 = \mu \times 15g\). Solving for \(\mu\), we get \(\mu = \frac{10}{15g} = 0.5\).

(i) To find the normal component of the force exerted on B by the ground, resolve the forces vertically. The weight of the block is \(7 \times 10 = 70\) N. The vertical component of the force \(X\) is \(X \cos 15^{\circ}\). In equilibrium, the normal force \(N\) plus the vertical component of \(X\) equals the weight of the block:

\(N + X \cos 15^{\circ} = 70\)

Thus, the normal component is:

\(N = 70 - X \cos 15^{\circ}\)

(ii) For limiting equilibrium, the frictional force \(F\) is equal to the maximum friction \(\mu N\), where \(\mu = 0.4\). The horizontal component of the force \(X\) is \(X \sin 15^{\circ}\). Therefore:

\(X \sin 15^{\circ} = 0.4 (70 - X \cos 15^{\circ})\)

Solving for \(X\):

\(X \sin 15^{\circ} = 28 - 0.4X \cos 15^{\circ}\)

\(X (\sin 15^{\circ} + 0.4 \cos 15^{\circ}) = 28\)

\(X = \frac{28}{\sin 15^{\circ} + 0.4 \cos 15^{\circ}}\)

Calculating gives \(X \approx 43.4\).