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Find the two values of θ for which \( \tan \frac{\theta}{2} = - \frac{1}{\sqrt{3}} \), where \( 0 \leq \theta \leq 4\pi \) - Leaving Cert Mathematics - Question 4 - 2020
Question 4
Find the two values of θ for which \( \tan \frac{\theta}{2} = - \frac{1}{\sqrt{3}} \), where \( 0 \leq \theta \leq 4\pi \). The diagram shows OAB, a sector of a c... show full transcript
Worked Solution & Example Answer:Find the two values of θ for which \( \tan \frac{\theta}{2} = - \frac{1}{\sqrt{3}} \), where \( 0 \leq \theta \leq 4\pi \) - Leaving Cert Mathematics - Question 4 - 2020
Step 1
Find the two values of θ for which tan(θ/2) = -1/√3
Answer
To solve for ( \theta ), we start with the equation:
[ \tan \frac{\theta}{2} = - \frac{1}{\sqrt{3}} ].
The reference angle for ( \tan ) is ( \frac{\pi}{6} ). Since the tangent is negative, ( \theta/2 ) must be in the second and fourth quadrants.
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Second Quadrant:
( \frac{\theta}{2} = \pi - \frac{\pi}{6} = \frac{5\pi}{6} )
Thus, ( \theta = 2 \times \frac{5\pi}{6} = \frac{5\pi}{3} ). -
Fourth Quadrant:
( \frac{\theta}{2} = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} )
Thus, ( \theta = 2 \times \frac{11\pi}{6} = \frac{11\pi}{3} ).
In the interval ( 0 \leq \theta \leq 4\pi ), the two values are ( \frac{5\pi}{3} ) and ( \frac{11\pi}{3} ).
Step 2
Find |BC|
Answer
Given that the area of the sector ( A_{OCA} ) is given by the formula:
[ A = \frac{1}{2} r^2 \theta ]
Substituting the values,
[ 21 = \frac{1}{2} \times 7^2 \times 1.2 ]
[ 21 = \frac{1}{2} \times 49 \times 1.2 ]
[ A_{OCA} = 29.4 ].
Now, to find |OC|:
[ |OC| = 7 \times \sin(1.2) ]
Using ( \sin(1.2) \approx 0.932 ):
[ |OC| = 7 \times 0.932 = 6.524 ].
Now to find |BC| with |AB| = 7 cm:
[ |AB| = 7 - |OC| = 7 - 6.524 = 0.476 ].
Thus, the distance |BC| is approximately ( 0.5 ) cm, when rounded into 1 decimal place.