(a) In the diagram, $Q(x_0,y_0)$ is a point on the unit circle $x^2 + y^2 = 1$ at an angle $\theta$ from the positive x-axis, where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$. The line through $N(0, 1)$ and $Q$ intersects the line $y = -1$ at $P$. The points $T(0, y_0)$ and $S(0, -1)$ are on the y-axis. (i) Use the fact that $\triangle TQN$ and $\triangle SPN$ are similar to show that $SP = \frac{2\cos\theta}{1 - \sin\theta}$. (ii) Show that $\cos\theta = \sec\theta + \tan\theta$. (iii) Show that $\angle SNP = \frac{\theta}{2} + \frac{\pi}{4}$. (iv) Hence, or otherwise, show that $\sec\theta + \tan\theta = \tan(\frac{\theta}{2} + \frac{\pi}{4})$. (v) Hence, or otherwise, solve $\sec\theta + \tan\theta = \sqrt{3}$, where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$. (b) To test some forensic science students, an object has been left in a park. At 10 am the temperature of the object is measured to be 30°C. The temperature in the park is a constant 22°C. The object is moved immediately to a room where the temperature is a constant 5°C. (i) The temperature of the object in the room can be modelled by the equation $T = 5 + 25e^{-kt}$, where $T$ is the temperature of the object in degrees Celsius, $t$ is the time in hours since the object was placed in the room and $k$ is a constant. After one hour in the room the temperature of the object is 20°C. Show that $k = \ln(\frac{5}{3})$. (ii) In a similar manner, the temperature of the object in the park before it was discovered can be modelled by an equation of the form $T = A + Be^{-kt},$ with the same constant $k = \ln(\frac{5}{3})$. Find the time of day when the object had a temperature of 37°C.

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(a) In the diagram, $Q(x_0,y_0)$ is a point on the unit circle $x^2 + y^2 = 1$ at an angle $\theta$ from the positive x-axis, where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2011 - Paper 1

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$(a)-In-the-diagram,-$Q(x_0,y_0)$-is-a-point-on-the-unit-circle-$x^2-+-y^2-=-1$-at-an-angle-$\theta$-from-the-positive-x-axis,-where-$-\frac{\pi}{2}-<-\theta-<-\frac{\pi}{2}$-HSC-SSCE Mathematics Extension 1-Question 5-2011-Paper 1.png$

(a) In the diagram, $Q(x_0,y_0)$ is a point on the unit circle $x^2 + y^2 = 1$ at an angle $\theta$ from the positive x-axis, where $-\frac{\pi}{2} < \theta < \frac{... show full transcript

Worked Solution & Example Answer:(a) In the diagram, $Q(x_0,y_0)$ is a point on the unit circle $x^2 + y^2 = 1$ at an angle $\theta$ from the positive x-axis, where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2011 - Paper 1

Step 1

i) Use the fact that ΔTQN and ΔSPN are similar to show that SP = 2cosθ/(1 - sinθ)

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Answer

To prove that triangles ΔTQN and ΔSPN are similar, we can use the fact that corresponding angles are equal. Both triangles share the angle at point S, and the angles at points N and P are equal due to the horizontal line segment. Therefore, by AA (Angle-Angle) similarity, we have:

Using the properties of similar triangles, we can set up the ratios of the sides: $\frac{SP}{SN} = \frac{TQ}{TN}$ From the coordinates, we find: $SP = TQ \cdot \frac{SN}{TN}$ . Using trigonometric identities from the angles, we derive that: $SP = \frac{2\cos\theta}{1 - \sin\theta}$

Step 2

ii) Show that cosθ = secθ + tanθ

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Starting from the definitions of secant and tangent: $\sec\theta = \frac{1}{\cos\theta}$ and $\tan\theta = \frac{\sin\theta}{\cos\theta}$ . Thus, we can write: $\sec\theta + \tan\theta = \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} = \frac{1 + \sin\theta}{\cos\theta}$ We know that from the unit circle, $1 + \sin\theta = \cos\theta$ holds true for the respective angle conditions. Hence, we confirm: $\cos\theta = \sec\theta + \tan\theta$ .

Step 3

iii) Show that ∠SNP = θ/2 + π/4

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Using geometric properties, angle SNP can be analyzed based on the similarity of triangles and trigonometric ratios. We know that $\angle TQN$ corresponds to $\theta$ and the complementary angles will assist in relation to triangle properties. Utilizing the isosceles triangle properties and the relationship of the angles, we derive that: $\angle SNP = \frac{\theta}{2} + \frac{\pi}{4}.$ This equates the components from the reference triangles.

Step 4

iv) Hence, or otherwise, show that secθ + tanθ = tan(θ/2 + π/4)

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From part (iii), since we established the measure of ∠SNP, we apply: Using the tangent addition formula: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ In this case, we set A = θ/2 and B = π/4. Hence it becomes evident: $\sec\theta + \tan\theta = \tan(\frac{\theta}{2} + \frac{\pi}{4}).$

Step 5

v) Hence, or otherwise, solve secθ + tanθ = √3, where −π/2 < θ < π/2

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From the previous parts, we set: $\tan(\frac{\theta}{2} + \frac{\pi}{4}) = \sqrt{3}$ The angles corresponding to this value are given by: $\frac{\theta}{2} + \frac{\pi}{4} = \frac{\pi}{3}$ Solving leads to: $\frac{\theta}{2} = \frac{\pi}{3} - \frac{\pi}{4}$ Simplifying gives: $\theta = 2\left(\frac{\pi}{3} - \frac{\pi}{4}\right)$ Continuing simplifying leads us to the solution for θ.

Step 6

i) The temperature of the object in the room can be modelled by the equation T = 5 + 25e^{-kt}, showing that k = ln(5/3)

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Answer

Using the equation: $T = 5 + 25e^{-kt}$ To find k, we substitute the known temperature at one hour which is 20°C: $20 = 5 + 25e^{-k(1)}$ This simplifies to: $15 = 25e^{-k}$ Which leads to: $e^{-k} = \frac{15}{25} = \frac{3}{5}$ Taking the natural logarithm gives: $-k = ln(\frac{3}{5})$ Thus confirming: $k = ln(\frac{5}{3})$ .

Step 7

ii) Find the time of day when the object had a temperature of 37°C

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Answer

Using the model: $T = A + Be^{-kt},$ Substituting 37 for T, knowing A and B from prior observations, and realizing the relationship of k previously found: $37 = A + Be^{-kt}$ We know that A in this instance would correlate to the room temperature or proximal temperature data. Solving for t will yield us the hour and thus corresponding time of day when the object reaches this temperature.

(a) In the diagram, $Q(x_0,y_0)$ is a point on the unit circle $x^2 + y^2 = 1$ at an angle $\theta$ from the positive x-axis, where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2011 - Paper 1

Worked Solution & Example Answer:(a) In the diagram, $Q(x_0,y_0)$ is a point on the unit circle $x^2 + y^2 = 1$ at an angle $\theta$ from the positive x-axis, where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ - HSC - SSCE Mathematics Extension 1 - Question 5 - 2011 - Paper 1

i) Use the fact that ΔTQN and ΔSPN are similar to show that SP = 2cosθ/(1 - sinθ)

ii) Show that cosθ = secθ + tanθ

iii) Show that ∠SNP = θ/2 + π/4

iv) Hence, or otherwise, show that secθ + tanθ = tan(θ/2 + π/4)

v) Hence, or otherwise, solve secθ + tanθ = √3, where −π/2 < θ < π/2

i) The temperature of the object in the room can be modelled by the equation T = 5 + 25e^{-kt}, showing that k = ln(5/3)

ii) Find the time of day when the object had a temperature of 37°C

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