precalc.pptx
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The document contains summaries of 4 assessment questions: 1) A reasoning question about finding the perimeter and area of a shaded sector of a circle. 2) A kinematics question about estimating the speed, distance traveled, and altitude of a plane based on angle of elevation measurements. 3) A graphing question about identifying the amplitude, period, and vertical shift of a cosine function graph. 4) A proof involving three mutually tangent circles, showing that the triangle formed by their centers is a right triangle and finding the area of the shaded region.
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Part A) Proof
Method One proves that the angle at the centre (Angle AOB) is twice the angle at circumference (Angle APB) by showing that triangles OBP and OAP are isosceles triangles, which allows calculating the angles.
Method Two extends a radius to form a diameter and shows that triangles OCA, OAP and OBP are isosceles triangles. It then uses properties of exterior angles to calculate that the angle at the centre (Angle AOB) is twice the angle at circumference (Angle APB). Both methods demonstrate the relationship between the angles at the centre and circumference of a circle.
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B. parabola
C. hyperbola
D. circle
Mark for review (Will be highlighted on the review page)
12. Where are the foci for the ellipse
A.
B.
C.
D.
13. An experimental model for a suspension bridge is built. In one section, cable runs from the top of one tower down to the roadway, just touching it there, and up again to the top of a second tower. The towers stand 60 inches apart. At a point between the towers and 18 inches along the road from the base of one tower, the cable is 1.44 inches above the roadway. Find the height of the towers.
A. 11 in
B. 9.5 in
C. 9 in
D. 8.5 in
14. Identify the following equation without applying a rotation of axes: x2 – 2xy – 2y2 – 3x + 3y + 10 = 0
A. circle
B. ellipse
C. hyperbola
D. parabola
15. Find the standard form of the equation of the ellipse satisfying the following conditions.Major axis vertical with length 18; length of minor axis = 12; center (0, 0)
A.
B.
C.
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