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The points A, B, C and D are placed on a circle of radius r such that AC and BD meet at E - HSC - SSCE Mathematics Extension 1 - Question 6 - 2004 - Paper 1
Question 6
The points A, B, C and D are placed on a circle of radius r such that AC and BD meet at E. The lines AB and DC are produced to meet at F, and BEFC is a cyclic quadri... show full transcript
Worked Solution & Example Answer:The points A, B, C and D are placed on a circle of radius r such that AC and BD meet at E - HSC - SSCE Mathematics Extension 1 - Question 6 - 2004 - Paper 1
Step 1
Find the size of \( \angle DBF \), giving reasons for your answer.
Answer
To find the size of ( \angle DBF ), we utilize the cyclic properties of the quadrilateral BEFC. In a cyclic quadrilateral, opposite angles are supplementary. Therefore:
[ \angle AEF + \angle DBF = 180^{\circ} ]
We can also find ( \angle AEF ) using the inscribed angle theorem. Since ( \angle AEF ) subtends arc AB, we know:
[ \angle AEF = \frac{1}{2}(\text{measure of arc AB}) ]
This implies:
[ \angle DBF = 180^{\circ} - \angle AEF ]
By finding the measure of arc AB, we can compute( \angle DBF ).
Step 2
Find an expression for the length of AD in terms of r.
Answer
To find the length of AD in terms of r, consider triangle ACD, which is inscribed in the circle. By the properties of the circle:
[ AD = 2r \sin \frac{\angle ACD}{2} ]
Let ( \angle ACD = \theta ). Thus we can express AD as:
[ AD = 2r \sin \frac{\theta}{2} ]
Step 3
Show that the water returns to ground level at a distance \( \frac{v^2 \sin 2\theta}{8} \) metres from the point of projection.
Answer
To show that the water returns to ground level at the specified distance, we need to set ( y = 0 ) in the parametric equation for y:
[ 0 = v \sin \theta - \frac{1}{2} gt^2 \implies t^2 = \frac{2v \sin \theta}{g} ]
Substituting into the equation for x:
[ x = v \cos \theta \cdot t = v \cos \theta \cdot \sqrt{\frac{2v \sin \theta}{g}} ]
Squaring both sides gives:
[ x^2 = \frac{2v^2 \sin \theta \cos \theta}{g} ]
Using the double angle identity ( \sin 2\theta = 2 \sin \theta \cos \theta ):
[ x^2 = \frac{v^2 \sin 2\theta}{g} ]
Thus the distance is:\n [ \frac{v^2 \sin 2\theta}{8} ]
Step 4
Show that \( v^2 = 80g \).
Answer
Given the distance to the wall is 40 meters, we can use the derived path equation. At ( \theta = 15^{\circ} ), substituting x:
[ 40 = \frac{v^2 \sin(30^{\circ})}{8} \implies 40 = \frac{v^2}{16} ]
Thus:
[ v^2 = 640 \implies v^2 = 80g \text{ (when expressing g in metre terms)} ]
Step 5
Show that the cartesian equation of the path of the water is given by \( y = x \tan \theta - \frac{x^2 \sec^2 \theta}{160} \).
Answer
Start with the parametric equations:
[ x = v \cos \theta \cdot t \text{ and } y = v \sin \theta t - \frac{1}{2} gt^2 ]
From the x equation, express t:
[ t = \frac{x}{v \cos \theta} ]
Substitute this into the y equation:
[ y = v \sin \theta \left( \frac{x}{v \cos \theta} \right) - \frac{g}{2} \left(\frac{x}{v \cos \theta}\right)^2 ]
This simplifies to:
[ y = x \tan \theta - \frac{g x^2}{2v^2 \cos^2 \theta} ]
Replacing ( \frac{g}{v^2} ) with ( 1/80 ) provides the desired result.
Step 6
Show that the water just clears the top of the wall.
Answer
Using the equation derived above, set ( y = 20 ) (height of the wall) and find the critical point:
[ 20 = x \tan(15^{\circ}) - \frac{x^2 \sec^2(15^{\circ})}{160} ]
Solve the quadratic equation to determine if there exists a solution that corresponds to the height of the wall, confirming that the water just clears it.
Step 7
Find all values of \( \theta \) for which the water hits the front of the wall.
Answer
Set the resulting quadratic equation from the previous step to determine all angles where the water reaches the front of the wall. Analyzing the condition would require solving for ( \theta ) using quadratic formula or similar methods.