A Math

A Math (IGCSE)

E Math

E Math (IGCSE)

H1 Math

H2 Math

IB AA HL Math

IB AA SL Math

IB AI HL Math

IB AI SL Math

IP Math

N(A) A Math

N(A) E Math

PSLE

PSLE Foundation Mathematics

<p>The diagram above shows two rods \(AB\) and \(BC\) of length \(4\) m and \(9\) m respectively and \(\angle ABC = 90^{\circ} \). Rollers are fixed at points \(A\) and \(B\) such that&nbsp;\(A\) is able to move along the wall and \(B\) is able to move along the celling. The horizontal distance of \(C\) from the vertical wall is \(d\) m.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1781508142282-maxm0gnyuj8.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODE1MDgxNDIyODItbWF4bTBnbnl1ajgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDg1MSwiZXhwIjoxNzkzNTA1MTUxfQ.rGGeCOgDgPWej65H40PSYAvUl1crD6ECM8v7QxytREs" style="max-width: min(320px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">Show clearly that&nbsp;\(d = 4\sin \theta + 9\cos \theta\).<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">Express \(d\) in the form \(R\cos(\theta - \alpha)\) where&nbsp;\(R &gt; 0\) and \(0^{\circ} &lt; \alpha &lt; 90^{\circ}\).<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-3" data-list-item-path="item-3"><div class="list-item-paragraph">Find the value of \(\theta\) for which \(d = 6\text{ m}\).<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-4" data-list-item-path="item-4"><div class="list-item-paragraph">Find the maximum value of \(d\) and the corresponding value of \(\theta\).<span class="inline-marks">[2]</span></div></li></ol>

The diagram above shows two rods \(AB\) and \(BC\) of length \(4\) m and \(9\) m respectively and \(\angle ABC = 90^{\circ} \). Rollers are fixed at points \(A\) and \(B\) such that \(A\) is able to move along the wall and \(B\) is able to move along the celling. The horizontal distance of \(C\) from the vertical wall is \(d\) m.

Trigonometric Modelling

Trigonometric Equations

R-Formula and Trigonometric Extrema

Applications of Trigonometric Ratios

2024 FMSS PRELIMS P2 Q9

Trigonometry

<p>In the diagram, it is given that&nbsp;\(AB=15\) cm, \(BC=40\) cm and \(\angle OCB=\theta\). The vertical line \(OC\) is perpendicular to \(OA\) and \(AB\) is perpendicular to \(BC\).</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1772960969441-r5lvjsp0wve.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzI5NjA5Njk0NDEtcjVsdmpzcDB3dmUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDg1MSwiZXhwIjoxNzkzNTA1MTUxfQ.ay7wK2tVqjwv84WzKPLNVN5tH5v7qicQCXGoDIzYvkA" style="max-width: min(400px, 100%); height: auto; display: block;" alt="Geometric diagram for trigonometry problem" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-a-2"><div class="list-item-content"><span class="list-item-text">Show that&nbsp;\(OC=15\sin\theta + 40\cos\theta\).</span><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b-2"><div class="list-item-content"><span class="list-item-text">Express \(OC\) in the form \(R\sin(\theta + \alpha)\), where&nbsp;\(R\) is a positive constant and \(\alpha\) is an acute angle in degrees.</span><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-c-2"><div class="list-item-content"><span class="list-item-text">Given that&nbsp;\(OC\) is at the maximum, determine the shortest distance of \(B\) to \(OC\).</span><span class="inline-marks">[3]</span></div></li></ol>

In the diagram, it is given that \(AB = 15\) cm, \(BC = 40\) cm and \(\angle OCB = \theta\). The vertical line \(OC\) is perpendicular to \(OA\) and \(AB\) is perpendicular to \(BC\).

Show that \(OC = 15\sin\theta + 40\cos\theta\).

Express \(OC\) in the form \(R\sin(\theta + \alpha)\), where \(R\) is a positive constant and \(\alpha\) is an acute angle in degrees.

Given that \(OC\) is at the maximum, determine the shortest distance of \(B\) to \(OC\).

2025 AHS PRELIMS P2 Q4

<p>In triangle \(ABC\), \(a\), \(b\) and \(c\) are the opposite sides of the interior angles \(A\), \(B\) and \(C\), respectively. It is known that&nbsp;\(a = \sqrt{3} \sin A\).</p>
<ol class="i-list"><li><div class="list-item-paragraph">Show that&nbsp;\(\frac{abc}{2c^2 + 2ab \cos C} \leq \frac{\sqrt{3} \sin C}{4 - 2 \cos C}.\)</div></li></ol>
<p>You may find the famous <strong>Inequality of Arithmetic and Geometric Means (AM-GM Inequality)</strong><br>useful: For any non-negative real numbers \(x_1\) and \(x_2\), it holds that \[AM = \frac{x_1 + x_2}{2} \geq \sqrt{x_1 x_2} = GM,\]where&nbsp;\(AM\) and \(GM\) denotes the arithmetic mean and geometric mean of \(x_1\) and \(x_2\), respectively.</p>
<ol class="i-list" start="2"><li><div class="list-item-content"><span class="list-item-text">Hence, or otherwise, show that the maximum value of \(\frac{abc}{2c^2 + 2ab \cos C}\) is \(\frac{1}{2}\) and this maximum value is achieved when&nbsp;\(\Delta ABC\) is an equilateral triangle.</span></div></li></ol>

In triangle \(ABC\), \(a\), \(b\) and \(c\) are the opposite sides of the interior angles \(A\), \(B\) and \(C\), respectively. It is known that \(a = \sqrt{3} \sin A\).

Show that \[\frac{abc}{2c^2 + 2ab \cos C} \leq \frac{\sqrt{3} \sin C}{4 - 2 \cos C}.\]

You may find the famous **Inequality of Arithmetic and Geometric Means (AM-GM Inequality)**
useful: For any non-negative real numbers \(x_1\) and \(x_2\), it holds that \[AM = \frac{x_1 + x_2}{2} \geq \sqrt{x_1 x_2} = GM,\]where \(AM\) and \(GM\) denotes the arithmetic mean and geometric mean of \(x_1\) and \(x_2\), respectively.

Hence, or otherwise, show that the maximum value of \(\frac{abc}{2c^2 + 2ab \cos C}\) is \(\frac{1}{2}\) and this maximum value is achieved when \(\Delta ABC\) is an equilateral triangle.

Sine Rule

Cosine Rule

Application of AM-GM Inequality in Trigonometry

<p>Express \(7 \sin x - 8 \cos x\) in the form of \(R \sin (x - \alpha)\), where&nbsp;\(p, R &gt; 0\) and \(0^\circ &lt; \alpha &lt; 90^\circ\).</p>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">Find the acute angle \(x\) for which \(7 \sin x - 8 \cos x = 3\).</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">Find the minimum value of \(7 \sin x - 8 \cos x\) and the corresponding value of \(x\), for&nbsp;\(0^\circ &lt; x &lt; 360^\circ\).</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">the maximum value of \(\frac{1}{(7 \sin x - 8 \cos x)^2 + 3}\) and the corresponding value of \(x\), for&nbsp;\(0^\circ &lt; x &lt; 360^\circ\).</span></div></li></ol>

Express \(7 \sin x - 8 \cos x\) in the form of \(R \sin (x - \alpha)\), where \(p, R > 0\) and \(0^\circ < \alpha < 90^\circ\).

Find the acute angle \(x\) for which \(7 \sin x - 8 \cos x = 3\).

Find the minimum value of \(7 \sin x - 8 \cos x\) and the corresponding value of \(x\), for \(0^\circ < x < 360^\circ\).

the maximum value of \(\frac{1}{(7 \sin x - 8 \cos x)^2 + 3}\) and the corresponding value of \(x\), for \(0^\circ < x < 360^\circ\).

<p>The diagram below shows the chords \(PR\) and \(QS\) that intersect at the point \(M\). Given that&nbsp;\(PQ = 7\) cm, \(RS = 24\) cm, the angle \(\angle RMQ = 90^\circ\) and the angle \(\angle PRS = \theta\), where&nbsp;\(\theta\) varies.</p>
<div class="graph-block" data-graph-config="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" data-width="167" data-height="169" data-proportional style="width:167px;height:169px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:167px;--graph-aspect:167 / 169;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Prove that&nbsp;\(PR = 24 \cos \theta + 7 \sin \theta\).</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Express \(PR\) in the form of \(R \cos (\theta - \alpha)\), where&nbsp;\(R &gt; 0\) and \(0^\circ &lt; \alpha &lt; 90^\circ\).</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">Find the maximum length of \(PR\) and the corresponding value of \(\theta\).</div></li></ol>

The diagram below shows the chords \(PR\) and \(QS\) that intersect at the point \(M\). Given that \(PQ = 7\) cm, \(RS = 24\) cm, the angle \(\angle RMQ = 90^\circ\) and the angle \(\angle PRS = \theta\), where \(\theta\) varies.

Prove that \(PR = 24 \cos \theta + 7 \sin \theta\).

Express \(PR\) in the form of \(R \cos (\theta - \alpha)\), where \(R > 0\) and \(0^\circ < \alpha < 90^\circ\).

Find the maximum length of \(PR\) and the corresponding value of \(\theta\).

<p>The diagram below shows a robotic arm \(PQR\) such that&nbsp;\(PQ\) makes an angle of \(\theta\) with the \(y\)-axis, where&nbsp;\(0^\circ &lt; \theta &lt; 90^\circ\) and the angle \(PQR = 90^\circ\). The lengths of \(PQ\) and \(QR\) are \(6\) m and \(4\) m respectively.</p>
<div class="graph-block" data-graph-config="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" data-width="227" data-height="256" data-proportional style="width:227px;height:256px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:227px;--graph-aspect:227 / 256;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Given that the vertical length of the robotic arm is \(L\) m,</div><ol class="abc-list"><li data-list-item-id="item-i-a" data-list-item-path="item-i/item-i-a"><div class="list-item-paragraph">show that&nbsp;\(L = 4 \sin \theta + 6 \cos \theta\), and</div></li><li data-list-item-id="item-i-b" data-list-item-path="item-i/item-i-b"><div class="list-item-paragraph">express \(L\) in the form \(R \sin (\theta + \alpha)\), where&nbsp;\(R &gt; 0\) and \(0^\circ &lt; \alpha &lt; 90^\circ\).</div></li></ol></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Based on your answers in <strong>(i)</strong>,</div><ol class="abc-list"><li data-list-item-id="item-ii-a" data-list-item-path="item-ii/item-ii-a"><div class="list-item-paragraph">find the maximum length of \(L\) and find the corresponding value of \(\theta\), and</div></li><li data-list-item-id="item-ii-b" data-list-item-path="item-ii/item-ii-b"><div class="list-item-paragraph">find the value(s) of \(\theta\) when&nbsp;\(L = 7\) m.</div></li></ol></li></ol>

The diagram below shows a robotic arm \(PQR\) such that \(PQ\) makes an angle of \(\theta\) with the \(y\)-axis, where \(0^\circ < \theta < 90^\circ\) and the angle \(PQR = 90^\circ\). The lengths of \(PQ\) and \(QR\) are \(6\) m and \(4\) m respectively.

Given that the vertical length of the robotic arm is \(L\) m,

<p>Express \(7 \cos 2x - 2 \sin 2x\) in the form of \(R \cos(2x + \alpha)\), where&nbsp;\(\alpha\) is in radians. Hence solve \(7 \cos 2x - 2 \sin 2x = 5\) for&nbsp;\(0 \leq x \leq 2\pi\).</p>

Express \(7 \cos 2x - 2 \sin 2x\) in the form of \(R \cos(2x + \alpha)\), where \(\alpha\) is in radians. Hence solve \(7 \cos 2x - 2 \sin 2x = 5\) for \(0 \leq x \leq 2\pi\).

<p>Express \(3 \sin x - 2 \cos x\) in the form of \(R \sin (x - \alpha)\), where&nbsp;\(\alpha\) is in radians. Hence solve \(3 \sin x - 2 \cos x = -2.5\) for&nbsp;\(0 \leq x \leq 2\pi\).</p>

Express \(3 \sin x - 2 \cos x\) in the form of \(R \sin (x - \alpha)\), where \(\alpha\) is in radians. Hence solve \(3 \sin x - 2 \cos x = -2.5\) for \(0 \leq x \leq 2\pi\).

<p>The diagram shows an L-shaped rod \(ABC\) leaning against a vertical wall. </p>
<p>\(AB=8\) cm, \(BC=15\) cm and angle \(\angle BCO=\theta\).</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768293828392-aywndi308zh.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgyOTM4MjgzOTItYXl3bmRpMzA4emgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDg1MSwiZXhwIjoxNzkzNTA1MTUxfQ.D6LFEZaKRR6AKohgcM4XjmeoQMTHH84OLK90ESqZ2O0" style="max-width: min(200px, 100%); height: auto; display: block;" alt="L-shaped rod ABC leaning against a wall" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Show that \(OC = 8 \sin \theta + 15 \cos \theta\).</span><span class="inline-marks">[2]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Express \(OC\) in the form \(R \sin(\theta + \alpha)\), where \(R &gt; 0\) and \(0{}^\circ &lt; \alpha &lt; 90{}^\circ\).</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the values of \(\theta\) when \(OC=16\) cm.</span><span class="inline-marks">[2]</span></div></li></ol>

The diagram shows an L-shaped rod \(ABC\) leaning against a vertical wall. 

\(AB=8\) cm, \(BC=15\) cm and angle \(\angle BCO=\theta\).

Show that \(OC = 8 \sin \theta + 15 \cos \theta\).

Express \(OC\) in the form \(R \sin(\theta + \alpha)\), where \(R > 0\) and \(0{}^\circ < \alpha < 90{}^\circ\).

Find the values of \(\theta\) when \(OC=16\) cm.

2023 ACSBR AM S3 EOY Q11 

<p>The diagram below shows the chords \(PQ\) and \(RS\) and they intersect at the point \(X\). Given that&nbsp;\(PQ=7\) cm, \(RS=24\) cm, \(\angle RXQ=90{}^\circ\) and \(\angle PRS=\theta\) where&nbsp;\(\theta\) varies.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2-1" data-layout="2-1"><div class="question-column" style="flex: 2; align-self: flex-start;"><ol class="abc-list"><li data-list-item-id="item-1780548663457-r781zxsue"><div class="list-item-paragraph">Prove that&nbsp;\(PR = 24 \cos \theta + 7 \sin \theta\).</div></li><li data-list-item-id="item-1780548664001-0ii36cfa5"><div class="list-item-paragraph">Express \(PR\) in the form \(R \cos(\theta - \alpha)\).</div></li><li data-list-item-id="item-1780548664352-k2zljpf4c"><div class="list-item-paragraph">Find the maximum length of \(PR\) and the corresponding value of \(\theta\).</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768102861843-0qiwbhfd6kk9.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDI4NjE4NDMtMHFpd2JoZmQ2a2s5LnBuZyIsInNjb3BlIjoiZG93bmxvYWQiLCJpYXQiOjE3OTM1MDQ4NTEsImV4cCI6MTc5MzUwNTE1MX0.EYqvisrIMoKcYguBWL8mcQF-ozzJqyQPBxpeCffuwcM" style="max-width: min(300px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

The diagram below shows the chords \(PQ\) and \(RS\) and they intersect at the point \(X\). Given that \(PQ=7\) cm, \(RS=24\) cm, \(\angle RXQ=90{}^\circ\) and \(\angle PRS=\theta\) where \(\theta\) varies.

<p>The diagram shows a trapezium \(ORST\) inscribed in a circle with centre \(O\) and diameter \(16\) cm. \(OR\) makes an angle \(\theta\), \(0{}^\circ &lt; \theta &lt; 90{}^\circ\), with the diameter of the circle. \(ST\) is perpendicular to both&nbsp;\(RS\) and \(OT\). The perimeter of the trapezium is \(P\) cm.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2-1" data-layout="2-1"><div class="question-column" style="flex: 2; align-self: flex-start;"><ol class="abc-list"><li data-list-item-id="item-1780548600520-ol4l6gd8k"><div class="list-item-paragraph">Show that&nbsp;\(P = 8 + 24 \cos \theta + 8 \sin \theta\).</div></li><li data-list-item-id="item-1780548600968-lprtp57hi"><div class="list-item-paragraph">Express \(24 \cos \theta + 8 \sin \theta\) in the form of \(R \cos(\theta - \alpha)\). Hence, find the maximum value of \(P\) and the corresponding value of \(\theta\) at which this value occurs.</div></li><li data-list-item-id="item-1780548601568-pmbljj84d"><div class="list-item-paragraph">Find the value of \(\theta\) when&nbsp;\(P = 30\) cm.</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768102718306-7df8srwyum5.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDI3MTgzMDYtN2RmOHNyd3l1bTUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDg1MSwiZXhwIjoxNzkzNTA1MTUxfQ.wGlwECn6fKWTgdFvtY_FuiZThQc9dzUBBtDbuM3wWqo" style="max-width: min(300px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

The diagram shows a trapezium \(ORST\) inscribed in a circle with centre \(O\) and diameter \(16\) cm. \(OR\) makes an angle \(\theta\), \(0{}^\circ < \theta < 90{}^\circ\), with the diameter of the circle. \(ST\) is perpendicular to both \(RS\) and \(OT\). The perimeter of the trapezium is \(P\) cm.

<p>The diagram shows a quadrilateral divided into two right-angled triangles and a rectangle.<br>Given that&nbsp;\(\angle ADC = \theta{}^\circ\), \(AB = 7\) cm, \(AD = 15\) cm and \(\angle BAD = \angle BCD = \angle AFB = 90{}^\circ\).</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2-1" data-layout="2-1"><div class="question-column" style="flex: 2; align-self: flex-start;"><ol class="abc-list"><li data-list-item-id="item-1780548515975-2vm0rfxc7"><div class="list-item-paragraph">Find the length of \(BC\) in the form of \(a \sin \theta - b \cos \theta\). Hence show that the perimeter, \(P\) cm, of the quadrilateral \(ABCD\) is given by \(P = 22 + 22 \sin \theta + 8 \cos \theta\).</div></li><li data-list-item-id="item-1780548516847-6b0z8who0"><div class="list-item-paragraph">Expressing \(P\) in the form \(P = 22 + R \sin(\theta + \alpha)\).</div></li><li data-list-item-id="item-1780548517335-yo4pr6mso"><div class="list-item-paragraph">Hence state the maximum value of the perimeter and state the corresponding value of \(\theta\).</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768102555320-swimt9mjcxh.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDI1NTUzMjAtc3dpbXQ5bWpjeGgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDg1MSwiZXhwIjoxNzkzNTA1MTUxfQ.DEVXXQVrMWjcqSDpzWB_oQP-NiYlgZPEspBu1vB0Th0" style="max-width: min(300px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

The diagram shows a quadrilateral divided into two right-angled triangles and a rectangle.
Given that \(\angle ADC = \theta{}^\circ\), \(AB = 7\) cm, \(AD = 15\) cm and \(\angle BAD = \angle BCD = \angle AFB = 90{}^\circ\).

<div class="content-block content-block--columns"><div class="question-columns question-columns--2-1" data-layout="2-1"><div class="question-column" style="flex: 2; align-self: flex-start;"><p>The diagram shows four points \(A\), \(B\), \(C\) and \(O\) such that that&nbsp;\(AB = 2\) m, \(BC = 10\) m,<br>\(\angle AOC = \angle ABC = 90{}^\circ\) and \(\angle OCB = \theta{}^\circ\).</p><ol class="abc-list"><li data-list-item-id="item-1780548397790-cpclwcpn6"><div class="list-item-paragraph">Show that&nbsp;\(OC = 2\sin \theta + 10\cos \theta\).</div></li><li data-list-item-id="item-1780548398221-uzovu47tj"><div class="list-item-paragraph">Express \(OC\) in the form \(R\sin(\theta + \alpha)\).</div></li><li data-list-item-id="item-1780548398613-t3y5ob0ft"><div class="list-item-paragraph">Find the maximum value of \(OC\) and the corresponding value of \(\theta\).</div></li><li data-list-item-id="item-1780548399229-kusfvjqx4"><div class="list-item-paragraph">Find the value of \(\theta\) for which \(OC = 8\) m.</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768102467338-ny3m0f7qie8.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDI0NjczMzgtbnkzbTBmN3FpZTgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDg1MSwiZXhwIjoxNzkzNTA1MTUxfQ.3MZ0yOSssJHrRGRxuVgkRkBFDIT-aqIr3Ge2tCfox_k" style="max-width: min(300px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

<div class="content-block content-block--columns"><div class="question-columns question-columns--2-1" data-layout="2-1"><div class="question-column" style="flex: 2; align-self: flex-start;"><p>The diagram shows an L-shaped dustpan \(ABC\) leaning against a vertical wall.<br>It is given that&nbsp;\(AB = 20\) cm, \(BC = 50\) cm, \(\angle AOC = \angle ABC = 90{}^\circ\) and \(\angle OCB = \theta{}^\circ\).</p><ol class="abc-list"><li data-list-item-id="item-1780548281620-y6yqd6q9k"><div class="list-item-paragraph">Show that&nbsp;\(AO + OC = (30 \cos \theta + 70 \sin \theta)\) cm.</div></li><li data-list-item-id="item-1780548282652-51t5lphme"><div class="list-item-paragraph">Find the value of \(\theta\) for which \(AO + OC = 65\) cm.</div></li><li data-list-item-id="item-1780548283244-2wzry829i"><div class="list-item-paragraph">State the maximum value of \((AO + OC)\) and the corresponding value of \(\theta\).</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768102193710-f44e9h7yijm.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDIxOTM3MTAtZjQ0ZTloN3lpam0ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MzUwNDg1MSwiZXhwIjoxNzkzNTA1MTUxfQ.ZGDzyfvdxwkr-ypJq2R1XKraWwryaU8Iosp_Z6q97VQ" style="max-width: min(300px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

<p>The diagram shows a robotic arm \(PQR\) such that&nbsp;\(PQ\) makes an angle of \(\theta\) with the \(y\)-axis, where&nbsp;\(0{}^\circ &lt; \theta &lt; 90{}^\circ\). \(QR\) makes an angle of \(\theta\) with the horizontal. The lengths of \(PQ\) and \(QR\) are \(6\) m and \(4\) m respectively.</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li data-list-item-id="item-1780548132946-8829rjc6m"><div class="list-item-paragraph">Given that the vertical length of the robotic arm is \(L\) m,</div><ol class="i-list"><li data-list-item-id="subitem-1780548136403-gr4kqztcb"><div class="list-item-paragraph">show that&nbsp;\(L=4\sin \theta + 6\cos \theta\).</div></li><li data-list-item-id="subitem-1780548137458-k0wv8ir0e"><div class="list-item-paragraph">express \(L\) in the form \(R\sin(\theta + \alpha){}^\circ\), where&nbsp;\(R&gt;0\) and \(0{}^\circ &lt; \theta &lt; 90{}^\circ\).</div></li></ol></li><li data-list-item-id="item-1780548134034-hl9nsfoyd"><div class="list-item-paragraph">Hence,</div><ol class="i-list"><li data-list-item-id="subitem-1780548144203-fiw6t1v8q"><div class="list-item-paragraph">state the maximum length of \(L\) and find the corresponding value of \(\theta\).</div></li><li data-list-item-id="subitem-1780548144610-ukwa0l0py"><div class="list-item-paragraph">find the values of \(\theta\) when&nbsp;\(L=7\) m.</div></li></ol></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768101995709-k9s03wd7r1.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDE5OTU3MDktazlzMDN3ZDdyMS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkzNTA0ODUxLCJleHAiOjE3OTM1MDUxNTF9.crGc8hIWOemZcYxFQnFJwnz9lH-Sv6RAqlWH5gNcE0A" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

The diagram shows a robotic arm \(PQR\) such that \(PQ\) makes an angle of \(\theta\) with the \(y\)-axis, where \(0{}^\circ < \theta < 90{}^\circ\). \(QR\) makes an angle of \(\theta\) with the horizontal. The lengths of \(PQ\) and \(QR\) are \(6\) m and \(4\) m respectively.

<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">Express \(5\) \(\sin 3x - 8\) \(\cos 3x\) in the form of \(R \sin(3x - \alpha)\) where&nbsp;\(\alpha\) is in radians.</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">Hence, solve \(5\) \(\sin 3x - 8\) \(\cos 3x = -6.5\) for&nbsp;\(0 \le x \le \frac{4}{3}\pi\).</span></div></li></ol>

Express \(5\) \(\sin 3x - 8\) \(\cos 3x\) in the form of \(R \sin(3x - \alpha)\) where \(\alpha\) is in radians.

Hence, solve \(5\) \(\sin 3x - 8\) \(\cos 3x = -6.5\) for \(0 \le x \le \frac{4}{3}\pi\).

<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">Express \(7\) \(\cos 2x - 2\) \(\sin 2x\) in the form of \(R \cos(2x + \alpha)\) where&nbsp;\(\alpha\) is in radians.</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">Hence, solve \(7\) \(\cos 2x - 2\) \(\sin 2x = 5\) for&nbsp;\(0 \le x \le 2\pi\).</span></div></li></ol>

Express \(7\) \(\cos 2x - 2\) \(\sin 2x\) in the form of \(R \cos(2x + \alpha)\) where \(\alpha\) is in radians.

Hence, solve \(7\) \(\cos 2x - 2\) \(\sin 2x = 5\) for \(0 \le x \le 2\pi\).

<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">Express \(\cos x - 6\) \(\sin x\) in the form of \(R \cos(x + \alpha)\) where&nbsp;\(\alpha\) is in radians.</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">Solve \(\cos x - 6\) \(\sin x = 2\) for&nbsp;\(0 \le x \le 2\pi\).</span></div></li></ol>

Express \(\cos x - 6\) \(\sin x\) in the form of \(R \cos(x + \alpha)\) where \(\alpha\) is in radians.

Solve \(\cos x - 6\) \(\sin x = 2\) for \(0 \le x \le 2\pi\).

<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">Express \(3\) \(\sin x - 2\) \(\cos x\) in the form of \(R \sin(x - \alpha)\) where&nbsp;\(\alpha\) is in radians.</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">Hence, solve \(3\) \(\sin x - 2\) \(\cos x = -2.5\) for&nbsp;\(0 \le x \le 2\pi\).</span></div></li></ol>

Express \(3\) \(\sin x - 2\) \(\cos x\) in the form of \(R \sin(x - \alpha)\) where \(\alpha\) is in radians.

Hence, solve \(3\) \(\sin x - 2\) \(\cos x = -2.5\) for \(0 \le x \le 2\pi\).

<p>Express \(9\) \(\cos \theta + 5\) \(\sin \theta\) in the form of \(R \sin(\theta + \alpha)\) where&nbsp;\(\alpha\) is in radians.</p>

Express \(9\) \(\cos \theta + 5\) \(\sin \theta\) in the form of \(R \sin(\theta + \alpha)\) where \(\alpha\) is in radians.

Math Question Bank