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Given: \( P = 119° \) and \( Q = 61° \) Determine: 3.1.1 \( \text{cosec P} \times \tan Q \) 3.1.2 \( \cos^{2}(P + 2Q) \) Given: \( \frac{1}{2} \tan \theta = 2 \), where \( \theta \in [0° ; 90°] \) Show, without the use of a calculator, that \( \sin^{2} \theta + \cos^{2} \theta = 1 \) Solve for \( x \): \( \sin x = \tan 318° \), where \( x \in [0° ; 360°] \) - NSC Technical Mathematics - Question 3 - 2024 - Paper 2
Question 3
![Given:--\(-P-=-119°-\)-and-\(-Q-=-61°-\)-Determine:--3.1.1-\(-\text{cosec-P}-\times-\tan-Q-\)--3.1.2-\(-\cos^{2}(P-+-2Q)-\)--Given:--\(-\frac{1}{2}-\tan-\theta-=-2-\),-where-\(-\theta-\in-[0°-;-90°]-\)-Show,-without-the-use-of-a-calculator,-that-\(-\sin^{2}-\theta-+-\cos^{2}-\theta-=-1-\)--Solve-for-\(-x-\):-\(-\sin-x-=-\tan-318°-\),-where-\(-x-\in-[0°-;-360°]-\)-NSC Technical Mathematics-Question 3-2024-Paper 2.png](https://cdn.simplestudy.cloud/assets/backend/uploads/question/subject_1520_paper_20926_q3_679a5deb.jpg)
Given: \( P = 119° \) and \( Q = 61° \) Determine: 3.1.1 \( \text{cosec P} \times \tan Q \) 3.1.2 \( \cos^{2}(P + 2Q) \) Given: \( \frac{1}{2} \tan \theta = 2 \... show full transcript
Worked Solution & Example Answer:Given: \( P = 119° \) and \( Q = 61° \) Determine: 3.1.1 \( \text{cosec P} \times \tan Q \) 3.1.2 \( \cos^{2}(P + 2Q) \) Given: \( \frac{1}{2} \tan \theta = 2 \), where \( \theta \in [0° ; 90°] \) Show, without the use of a calculator, that \( \sin^{2} \theta + \cos^{2} \theta = 1 \) Solve for \( x \): \( \sin x = \tan 318° \), where \( x \in [0° ; 360°] \) - NSC Technical Mathematics - Question 3 - 2024 - Paper 2
Step 1
Determine: 3.1.1 cosec P × tan Q
Answer
To find ( \csc P \times \tan Q ), we first need to calculate ( \csc P ) and ( \tan Q ):
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Calculate ( \csc P ): [ \csc P = \frac{1}{\sin P} = \frac{1}{\sin(119°)} \approx 1.06 ]
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Calculate ( \tan Q ): [ \tan Q = \tan(61°) \approx 1.80 ]
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Combine the results: [ \csc P \times \tan Q = 1.06 \times 1.80 \approx 1.908 ] So, ( \csc P \times \tan Q \approx 2.06 ).
Step 2
Determine: 3.1.2 cos²(P + 2Q)
Answer
To find ( \cos^{2}(P + 2Q) ), we can follow these steps:
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Calculate ( P + 2Q ): [ P + 2Q = 119° + 2 \times 61° = 119° + 122° = 241° ]
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Calculate ( \cos(241°) ): [ \cos(241°) = -0.24 ] (Note: Since 241° is in the third quadrant, cosine is negative.)
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Calculate ( \cos^{2}(P + 2Q) ): [ \cos^{2}(241°) = (-0.24)^{2} = 0.0576 \approx 0.24 ].
Step 3
Show, without the use of a calculator, that \( \sin^{2} \theta + \cos^{2} \theta = 1 \)
Answer
Given: ( \frac{1}{2} \tan \theta = 2 ), we first find ( \tan \theta = 4 ).
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Determine the relationship of sine and cosine: We know that: [ \tan \theta = \frac{\sin \theta}{\cos \theta} ] Then substituting: [ \tan \theta = 4 = \frac{\sin \theta}{\cos \theta} \implies \sin \theta = 4 \cos \theta ]
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Substitute into sin² + cos²: [ \sin^{2} \theta + \cos^{2} \theta = (4 \cos \theta)^{2} + \cos^{2} \theta = 16 \cos^{2} \theta + \cos^{2} \theta = 17 \cos^{2} \theta ] We break it down as: [ \frac{17 \cos^{2} \theta}{17} = 1 \quad \text{using the Pythagorean identity.} ] Therefore, it confirms that ( \sin^{2} \theta + \cos^{2} \theta = 1 ).
Step 4
Solve for x: 3.3 sin x = tan 318°
Answer
To solve ( \sin x = \tan 318° ), we first compute ( \tan 318° ):
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Calculate ( \tan 318° ): Knowing ( 318° = 360° - 42°): [ \tan 318° = -\tan 42°\approx -0.9004 ]
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Set the equation up: [ \sin x = -0.9004 ] This means we look for angles in the third and fourth quadrants:
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Find reference angle and solve for x: The reference angle is 42°, giving us:
- In the third quadrant: ( x = 180° + 42° = 222° )
- In the fourth quadrant: ( x = 360° - 42° = 318° )
Thus, the solutions for ( x ) are approximately ( 222.21° ) and ( 318° ).