Elementary Mathematics 2016 (II)

Step-by-step solution

1. 1
   Place $B=(0,0)$, $C=(b+c,0)$, $A=(b,a)$

   Why: Take the side $(b+c)$ as the base and look for a point whose distances to its two ends are the other two given lengths.

2. 2
   $AB=\sqrt{b^2+a^2}$ and $AC=\sqrt{c^2+a^2}$

   Why: The horizontal offsets from $A$ to $B$ and to $C$ are $b$ and $c$, and the vertical offset is $a$ in both cases.

   Pythagoras' theorem
3. 3
   So the triangle has base $b+c$ and height $a$.

   Why: These are exactly the three given side lengths, so this placement is the triangle described; the foot of the altitude from $A$ splits the base into $b$ and $c$.

4. 4
   $\text{Area}=\frac{1}{2}(b+c)(a)=\frac{a(b+c)}{2}$

   Why: Half base times height.

   Area of a triangle from base and height

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Step-by-step solution

1. 1
   $\text{base}=2r$

   Why: The diameter is the longest chord of the semicircle, so no other choice of base is larger.

2. 2
   $\text{greatest height above the diameter}=r$

   Why: Every point of the arc is $r$ from the centre, so its distance above the diameter is at most $r$, attained at the midpoint of the arc.

3. 3
   $\text{Area}=\frac{1}{2}(2r)(r)=r^2$ square units

   Why: Half base times height, with both factors at their maxima simultaneously, so this really is the largest such triangle.

   Area of a triangle from base and height

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Step-by-step solution

1. 1
   $C_{\text{front}}=\pi (2x)=2\pi x$, $C_{\text{rear}}=\pi (2y)=2\pi y$

   Why: Circumference is $\pi$ times the diameter, and the diameters are given as $2x$ and $2y$.

   Area and circumference of a circleDistance covered by a rolling wheel
2. 2
   $\text{distance}=n\cdot 2\pi x$

   Why: Each revolution of the front wheel advances the engine by one front circumference.

3. 3
   $N=\frac{2\pi nx}{2\pi y}=\frac{nx}{y}$

   Why: The rear wheel covers the same distance, so its revolution count is that distance divided by its own circumference; $2\pi$ cancels, so the answer depends only on the diameters.

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Step-by-step solution

1. 1
   $r=\frac{126}{2}=63$ m

   Why: The cone sits on the cylinder, so they share the base radius; the stem gives the diameter.

2. 2
   $h_{\text{cone}}=21-5=16$ m

   Why: The $21$ m is the total height of the tent, of which the cylindrical part accounts for $5$ m.

3. 3
   $l=\sqrt{63^2+16^2}=\sqrt{3969+256}=\sqrt{4225}=65$ m

   Why: The slant height is the hypotenuse of the right triangle formed by the cone's vertical height and its base radius.

   Pythagoras' theorem

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Step-by-step solution

1. 1
   $r=\frac{126}{2}=63$ m, $h=5$ m

   Why: The stem gives the cylindrical part's diameter and height directly.

2. 2
   $\text{CSA}=2\pi rh=2\times \frac{22}{7}\times 63\times 5$

   Why: The curved surface of a cylinder is the base circumference times the height.

   Curved and total surface of a cylinder
3. 3
   $=2\times 22\times 9\times 5=1980$ m${}^2$

   Why: Cancel $\frac{63}{7}=9$ before multiplying so the arithmetic stays in two digits.

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Step-by-step solution

1. 1
   $\text{cylinder CSA}=2\pi rh=2\times \frac{22}{7}\times 63\times 5=1980$ m${}^2$

   Why: The cylindrical wall of the tent, from the stem's $r=63$ m and $h=5$ m.

   Curved and total surface of a cylinder
2. 2
   $l=\sqrt{63^2+16^2}=65$ m

   Why: The cone's height is $21-5=16$ m, and the slant height is needed before its curved surface can be found.

   Pythagoras' theorem
3. 3
   $\text{cone CSA}=\pi rl=\frac{22}{7}\times 63\times 65=198\times 65=12870$ m${}^2$

   Why: The conical roof; $\frac{22\times 63}{7}=198$ keeps the multiplication small.

   Slant height and curved surface of a cone
4. 4
   $\text{canvas}=1980+12870=14850$ m${}^2$

   Why: Only the two curved surfaces are made of canvas: the tent has no canvas floor, and the circle where cone meets cylinder is an internal join, not a surface.

   Volume and surface of a composite solid

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Step-by-step solution

1. 1
   $\text{distance}=66\times \frac{10}{60}=11$ km

   Why: Ten minutes is $\frac{1}{6}$ of an hour, and distance $=$ speed $\times$ time.

   Speed, distance and time
2. 2
   $11\text{ km}=1100000$ cm

   Why: The wheel's diameter is in centimetres, so the distance must be too: $1$ km $=100000$ cm.

3. 3
   $C=\pi d=\frac{22}{7}\times 80=\frac{1760}{7}$ cm

   Why: One complete revolution advances the car by one circumference.

   Area and circumference of a circleDistance covered by a rolling wheel
4. 4
   $N=\frac{1100000\times 7}{1760}=4375$

   Why: Divide the distance by the per-revolution distance; it comes out whole, so $4375$ revolutions are complete.

   Unitary method: find one unit, then scale

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1. 1
   $5^2+12^2=25+144=169=13^2$

   Why: The converse of Pythagoras: the squares on the two shorter sides sum to the square on the longest, so the triangle is right-angled with $13$ cm as the hypotenuse.

   Pythagoras' theorem
2. 2
   $\text{Area}=\frac{1}{2}\times 5\times 12=30$ cm${}^2$

   Why: The two legs are perpendicular, so they serve directly as base and height and Heron's formula is not needed.

   Area of a triangle from base and height

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Step-by-step solution

1. 1
   $2\pi r=4a=P$

   Why: The circle's circumference and the square's perimeter are the same quantity $P$.

   Area and circumference of a circlePerimeter and area of a square
2. 2
   $r=\frac{P}{2\pi },a=\frac{P}{4}$

   Why: Solve each perimeter relation for its own length.

3. 3
   $\frac{\pi r^2}{a^2}=\frac{P^2/(4\pi )}{P^2/16}=\frac{16}{4\pi }=\frac{4}{\pi }$

   Why: $P^2$ cancels, so the ratio does not depend on how big the figures are.

4. 4
   $\frac{4}{\pi }=4\times \frac{7}{22}=\frac{28}{22}=\frac{14}{11}$, so the ratio is $14:11$

   Why: The printed options are in integers, which forces $\pi =\frac{22}{7}$.

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Step-by-step solution

1. 1
   $AB=2r$ and $CD=\frac{1}{2}AB=r$

   Why: $AB$ is a diameter, and the stem says $CD$ is half of it.

2. 2
   Let $M$ be the midpoint of $CD$; then $OM\perp CD$ and $MC=\frac{r}{2}$

   Why: The segment from the centre to the midpoint of a chord is perpendicular to that chord.

   The perpendicular from the centre bisects a chord
3. 3
   $O{M}^2+{\left(\frac{r}{2}\right)}^2=r^2\Rightarrow OM=\frac{\sqrt{3}r}{2}$

   Why: $OC$ is a radius, so $△OMC$ is right-angled at $M$ with hypotenuse $r$.

   Pythagoras' theorem
4. 4
   $M$ traces a circle centred at $O$ of radius $\frac{\sqrt{3}r}{2}$

   Why: Rotating the figure about $O$ preserves every distance from $O$, so $M$ traces the circle of radius $OM$.

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Step-by-step solution

1. 1
   $(\sin \theta +\cos \theta {)}^2={\left(\frac{\sqrt{7}}{2}\right)}^2=\frac{7}{4}$

   Why: Squaring is the one move that connects a sum of the two ratios to the identity relating their squares.

   Square of a sum and of a difference
2. 2
   $1+2\sin \theta \cos \theta =\frac{7}{4}\Rightarrow 2\sin \theta \cos \theta =\frac{3}{4}$

   Why: ${\sin }^2\theta +{\cos }^2\theta =1$ collapses the two square terms to $1$, leaving only the cross term.

   Pythagorean trigonometric identities
3. 3
   $(\sin \theta -\cos \theta {)}^2=1-2\sin \theta \cos \theta =1-\frac{3}{4}=\frac{1}{4}$

   Why: The difference has the same expansion with the cross term negated, so the quantity just found serves a second time.

4. 4
   $\sin \theta -\cos \theta =\pm \frac{1}{2}$

   Why: The square root carries a sign that the question does not fix; only the positive value is printed among the options.

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Step-by-step solution

1. 1
   $\sin x+{\sin }^{2}x=1\Rightarrow \sin x=1-{\sin }^{2}x={\cos }^{2}x$

   Why: $1-{\sin }^{2}x$ is ${\cos }^{2}x$, so the condition says exactly that $\sin x$ and ${\cos }^{2}x$ are the same number.

   Pythagorean trigonometric identities
2. 2
   ${\cos }^{8}x+2{\cos }^{6}x+{\cos }^{4}x={\cos }^{4}x\left({\cos }^{4}x+2{\cos }^{2}x+1\right)={\cos }^{4}x{\left({\cos }^{2}x+1\right)}^2$

   Why: Taking out ${\cos }^{4}x$ leaves a perfect square in ${\cos }^{2}x$.

   Square of a sum and of a difference
3. 3
   ${\cos }^{4}x{\left({\cos }^{2}x+1\right)}^2={\left[{\cos }^{2}x\left({\cos }^{2}x+1\right)\right]}^2$

   Why: Both factors are squares, so the whole expression is the square of a single product.

4. 4
   ${\cos }^{2}x\left({\cos }^{2}x+1\right)=\sin x\left(\sin x+1\right)=\sin x+{\sin }^{2}x=1$

   Why: Substituting ${\cos }^{2}x=\sin x$ reproduces the left-hand side of the given equation, which is $1$.

5. 5
   $\text{Value}=1^2=1$

   Why: The bracket is $1$, so its square is $1$.

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Step-by-step solution

1. 1
   $\tan 30^{\circ }=\frac{100}{d_1}\Rightarrow d_1=\frac{100}{\tan 30^{\circ }}=100\sqrt{3}$

   Why: The angle of depression from the top equals the angle of elevation from the car, so the tower's height over the far distance is $\tan 30^{\circ }$.

   Angles of elevation and depressionSine, cosine and tangent in a right triangleExact trigonometric ratios of the standard angles
2. 2
   $\tan 60^{\circ }=\frac{100}{d_2}\Rightarrow d_2=\frac{100}{\tan 60^{\circ }}=\frac{100}{\sqrt{3}}$

   Why: The same right triangle with the later, steeper angle gives the nearer distance.

3. 3
   $d_1-d_2=100\sqrt{3}-\frac{100}{\sqrt{3}}=\frac{300-100}{\sqrt{3}}=\frac{200}{\sqrt{3}}$

   Why: The car moved from the first position to the second, so what it travelled is the difference of the two distances.

4. 4
   $\frac{200}{\sqrt{3}}=\frac{200\sqrt{3}}{3}$

   Why: Rationalising puts the answer in the form the options are printed in.

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Step-by-step solution

1. 1
   $\tan 30^{\circ }=\frac{75}{d_1}\Rightarrow d_1=\frac{75}{\tan 30^{\circ }}=75\sqrt{3}$

   Why: The man who sees the smaller angle of elevation is the further away.

   Angles of elevation and depressionSine, cosine and tangent in a right triangleExact trigonometric ratios of the standard angles
2. 2
   $\tan 60^{\circ }=\frac{75}{d_2}\Rightarrow d_2=\frac{75}{\sqrt{3}}=25\sqrt{3}$

   Why: The same tower, the other observer, the steeper angle.

3. 3
   $\text{The men are on either side}\Rightarrow d=d_1+d_2$

   Why: Opposite sides put the foot of the tower between them, so the two distances add; on the same side they would subtract.

4. 4
   $d=75\sqrt{3}+25\sqrt{3}=100\sqrt{3}\text{m}$

   Why: Both terms carry $\sqrt{3}$, so they combine directly.

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Step-by-step solution

1. 1
   $\sec 56^{\circ }=\mathrm{cosec}\left(90^{\circ }-56^{\circ }\right)=\mathrm{cosec}34^{\circ }$

   Why: $56^{\circ }$ and $34^{\circ }$ are complementary, and the secant of an angle is the cosecant of its complement. This is what makes the second and third terms share an angle.

   Ratios of complementary angles
2. 2
   ${\sec }^{2}56^{\circ }-{\cot }^{2}34^{\circ }={\mathrm{cosec}}^{2}34^{\circ }-{\cot }^{2}34^{\circ }=1$

   Why: $1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta$ is one of the Pythagorean identities.

   Pythagorean trigonometric identities
3. 3
   $\tan 22^{\circ }=\cot \left(90^{\circ }-22^{\circ }\right)=\cot 68^{\circ }$

   Why: $22^{\circ }$ and $68^{\circ }$ are complementary, so the remaining two terms can also be brought to one angle.

4. 4
   ${\mathrm{cosec}}^{2}68^{\circ }-{\tan }^{2}22^{\circ }={\mathrm{cosec}}^{2}68^{\circ }-{\cot }^{2}68^{\circ }=1$

   Why: The same identity, now at $68^{\circ }$.

5. 5
   ${\mathrm{cosec}}^{2}68^{\circ }+{\sec }^{2}56^{\circ }-{\cot }^{2}34^{\circ }-{\tan }^{2}22^{\circ }=1+1=2$

   Why: The expression is exactly the sum of the two pairs.

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Step-by-step solution

1. 1
   $2y\cos \theta =x\sin \theta$

   Why: The first given, held ready to be substituted.

2. 2
   $2x\sec \theta -y\mathrm{cosec}\theta =3\Rightarrow 2x\sin \theta -y\cos \theta =3\sin \theta \cos \theta$

   Why: Multiplying through by $\sin \theta \cos \theta$ clears both reciprocal ratios; neither factor is zero, since $\sec \theta$ and $\mathrm{cosec}\theta$ are written down and so are defined.

   Reciprocal and quotient relations between the ratios
3. 3
   $2x\sin \theta =2\left(2y\cos \theta \right)=4y\cos \theta \Rightarrow 4y\cos \theta -y\cos \theta =3\sin \theta \cos \theta$

   Why: Doubling the first relation replaces $2x\sin \theta$, and the left-hand side collapses to a single term.

4. 4
   $3y\cos \theta =3\sin \theta \cos \theta \Rightarrow y=\sin \theta$

   Why: Dividing by $3\cos \theta$, which is non-zero.

5. 5
   $x\sin \theta =2y\cos \theta =2\sin \theta \cos \theta \Rightarrow x=2\cos \theta$

   Why: Putting $y=\sin \theta$ back into the first relation and cancelling the non-zero $\sin \theta$.

6. 6
   $x^2+4y^2=4{\cos }^2\theta +4{\sin }^2\theta =4\left({\sin }^2\theta +{\cos }^2\theta \right)=4$

   Why: The Pythagorean identity removes $\theta$ altogether, which is why the question has a numerical answer at all.

   Pythagorean trigonometric identities

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Step-by-step solution

1. 1
   ${\left(\sin \theta +\cos \theta \right)}^2={\left(\frac{1+\sqrt{3}}{2}\right)}^2=\frac{4+2\sqrt{3}}{4}=\frac{2+\sqrt{3}}{2}$

   Why: Squaring turns the given sum into something the Pythagorean identity can act on.

   Square of a sum and of a difference
2. 2
   $1+2\sin \theta \cos \theta =\frac{2+\sqrt{3}}{2}\Rightarrow \sin \theta \cos \theta =\frac{\sqrt{3}}{4}$

   Why: ${\sin }^2\theta +{\cos }^2\theta =1$ removes the square terms, leaving the product alone.

   Pythagorean trigonometric identities
3. 3
   $\tan \theta +\cot \theta =\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }=\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }=\frac{1}{\sin \theta \cos \theta }$

   Why: Adding the two over a common denominator makes the identity appear again in the numerator, so the answer depends on the product only - which is exactly what step 2 supplies.

4. 4
   $\tan \theta +\cot \theta =\frac{1}{\frac{\sqrt{3}}{4}}=\frac{4}{\sqrt{3}}$

   Why: Substituting the product found in step 2.

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Step-by-step solution

1. 1
   $t={\cos }^2\theta ,0\le \theta <\frac{\pi }{2}\Rightarrow 0<t\le 1$

   Why: ${\cos }^2\theta$ is $1$ at $\theta =0$ and falls towards $0$, which it never reaches because $\theta =\frac{\pi }{2}$ is excluded.

2. 2
   $A={\sin }^2\theta +{\cos }^4\theta =\left(1-t\right)+t^2=t^2-t+1$

   Why: ${\sin }^2\theta =1-{\cos }^2\theta$ puts the whole expression in one variable.

   Pythagorean trigonometric identities
3. 3
   $A={\left(t-\frac{1}{2}\right)}^2+\frac{3}{4}$

   Why: Completing the square exposes both the least value and the $t$ at which it occurs.

   Range of a quadratic by completing the square
4. 4
   $A_{\min }=\frac{3}{4}\text{at}t=\frac{1}{2},\text{i.e.}\theta =45^{\circ }$

   Why: The square is smallest when it is zero, and $t=\frac{1}{2}$ does lie in the allowed range, so this minimum is attained.

5. 5
   $A(1)=1\text{and}A(t)\to 1\text{as}t\to 0^{+}\Rightarrow A_{\max }=1$

   Why: An upward parabola is largest at an endpoint of the interval; $t=1$ is attained (at $\theta =0$) and gives $1$, while the other endpoint is only approached.

6. 6
   $\frac{3}{4}\le A\le 1$

   Why: Both bounds are reached, so the inequalities are not strict.