Question: [Maximum mark: 6]
The following diagram shows a circle with centre O and radius 5 metres.
Points A and B lie on the circle and \(A\hat{O}B\) 1.9 = radians.
(a) Find the length of the chord [AB].
(b) Find the area of the shaded sector.
Answer/Explanation
Ans:
(a) EITHER
uses the cosine rule
AB2 = 52 + 52 -2 × 5 × 5 × cos1.9
OR
uses right-angled trigonometry
\(\frac{\frac{AB}{2}}{5}= sin 0.95\)
OR
uses the sine rule
\(\alpha = \frac{1}{2}\left ( \pi -1.9 \right )\left ( =0.6207… \right )\)
\(\frac{AB}{sin 1.9}=\frac{5}{sin 0.6207…}\)
THEN
AB = 8.13415…
AB = 8.13 (m)
(b) let the shaded area be A
METHOD 1
attempt at finding reflex angle
METHOD 2
let the area of the circle be AC and the area of the unshaded sector be AU
A = AC – AU
\(A = \pi \times 5^{2}-\frac{1}{2}\times 5^{2}\times 1.9(78.5398…-23.75)\)
= 54.7898…
= 54.8 (m2)
Question
Two straight fences meet at point A and a field lies between them.
A horse is tied to a post, P, by a rope of length r metres. Point D is on one fence and
point E is on the other, such that PD = PE = PA = r and \(D\hat{P}E\) = θ radians. This is shown in
the following diagram.
The length of the arc DE shown in the diagram is 28 m.
Write down an expression for r in terms of θ . [1]
Show that the area of the field that the horse can reach is \(\frac{392}{\Theta ^{2}}(\Theta \;+\; Sin\Theta)\) [4]
The area of field that the horse can reach is 460 m2. Find the value of θ . [2]
Hence, find the size of \(D\hat{A}E\) . [2]
A new fence is to be constructed between points B and C which will enclose the field, as shown in the following diagram.
-
Point C is due west of B and AC = 800 m . The bearing of B from A is 195°.
e. (i) Find the size of \(A\hat{B}C\) .
(ii) Find the length of new fence required. [5]
Answer/Explanation
Ans:
(a) r= \(\frac{28}{\Theta }\)
(b)
recognising sum of area of sector and area of triangle required
\(\frac{1}{2}r^{2}\Theta+ \frac{1}{2}r \times r \times sin (\pi -\Theta )= \frac{r^{2}}{2}(\Theta +sin(\pi -\Theta )\)
\(sin (\pi -\Theta )= sin \Theta \) (substitution seen anywhere)
\(\frac{1}{2}(\frac{28}{\Theta})^{2}\Theta +\frac{1}{2}(\frac{28}{\Theta })^{2}sin \Theta\)
OR
\(\frac{1}{2}(\frac{28}{\Theta })^{2}(\Theta + Sin \Theta)\)
area = \(\frac{392}{\Theta ^{2}}(\Theta + sin \Theta )\)
(c)
\(\frac{392}{\Theta ^{2}}(\Theta + sin \Theta )= 460\)
\(\Theta = 1.43917..\)
\(\Theta = 1.44\)
(d)
\(\frac{\pi -(\pi -\Theta )}{2}\)
OR
\(\frac{\Theta }{2}\)
\(D\hat{A}E = 0.719588..\)
\(D\hat{A}E= 0.720\)
(e)
(i)
\(A\hat{B}C\) = 195 – 180 +90 = 1050
(ii)
choosing sine rule
\(\frac{BC}{sinD\hat{A}E }= \frac{800}{sin105 }\)
OR
\(\frac{BC}{sinD\hat{A}E}=\frac{800}{sin 1.83}\)
\(BC = 546(m)\)
Question
The following diagram shows a circle with centre O and radius 1 cm. Points A and B lie on the circumference of the circle and AÔB =2θ , where 0 < θ < \(\frac{\pi }{2}\) .
The tangents to the circle at A and B intersect at point C
Show that AC = tan θ . [1]
Find the value of θ when the area of the shaded region is equal to the area of sector OADB. [6]
Answer/Explanation
Ans
Question
[Maximum mark: 6] [without GDC]
(a) Express the following angles in radians; give your answer in terms of π.
(i) 20° (ii) 18° (iii) 540° [3]
(b) Express the following angles in degrees
(i) \(\frac{\pi}{18} \mathrm{rad}\) (ii) \(\frac{\pi}{5} \mathrm{rad}\) (iii) \(2.5 \pi\) [3]
Answer/Explanation
Ans:
(a) (i) $\frac{\pi}{9}$
(ii) $\frac{\pi}{10}$
(iii) $3 \pi$
(b) (i) 10° (ii) 36° (iii) 450°.
Question
[Maximum mark: 6] [without GDC]
The diagrams show a circular sector of radius 10 cm and angle θ radians which is formed into a cone of slant height 10 cm. The vertical height h of the cone is equal to the radius r of its base.
(a) Find the value of r . [2]
(b) Find the angle θ radians. [4]
Answer/Explanation
Ans:
\(h=r \text { so } 2 r^{2}=100 \Rightarrow r^{2}=50 \Rightarrow r=5 \sqrt{2} \quad \text { Hence circumference }=2 \pi r=10 \pi \sqrt{2}\)
\(l=10 \theta=10 \pi \sqrt{2} \Rightarrow \theta=\pi \sqrt{2}\)