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In \( \Delta MNP \), \( \hat{N} = 90^{\circ} \) and \( \sin M = \frac{15}{17} \), determine, without using a calculator: 5.1.1 tan M 5.1.2 The length of NP if MP = 51 5.2 Simplify to a single term: \( \cos(x - 360^\circ) \sin(90^\circ - x) + \cos^2(z - x) - 1 \) 5.3 Consider: \( \sin(2x + 40^\circ) \cos(x + 30^\circ) - \cos(2x + 40^\circ) \sin(2x + 30^\circ) \) 5.3.1 Write as a single trigonometric term in its simplest form - NSC Mathematics - Question 5 - 2018 - Paper 2
Question 5
In \( \Delta MNP \), \( \hat{N} = 90^{\circ} \) and \( \sin M = \frac{15}{17} \), determine, without using a calculator: 5.1.1 tan M 5.1.2 The length of NP if MP =... show full transcript
Worked Solution & Example Answer:In \( \Delta MNP \), \( \hat{N} = 90^{\circ} \) and \( \sin M = \frac{15}{17} \), determine, without using a calculator: 5.1.1 tan M 5.1.2 The length of NP if MP = 51 5.2 Simplify to a single term: \( \cos(x - 360^\circ) \sin(90^\circ - x) + \cos^2(z - x) - 1 \) 5.3 Consider: \( \sin(2x + 40^\circ) \cos(x + 30^\circ) - \cos(2x + 40^\circ) \sin(2x + 30^\circ) \) 5.3.1 Write as a single trigonometric term in its simplest form - NSC Mathematics - Question 5 - 2018 - Paper 2
Step 1
5.1.1 tan M
Answer
To determine ( \tan M ), we can use the relationship of sine and cosine in a right triangle. Since ( \sin M = \frac{15}{17} ), we can find ( \cos M ) using the Pythagorean theorem:
- Calculate ( \cos M = \sqrt{1 - \sin^2 M} = \sqrt{1 - \left(\frac{15}{17}\right)^2} = \sqrt{1 - \frac{225}{289}} = \sqrt{\frac{64}{289}} = \frac{8}{17} ).
- Therefore, ( \tan M = \frac{\sin M}{\cos M} = \frac{\frac{15}{17}}{\frac{8}{17}} = \frac{15}{8} ).
Step 2
5.1.2 The length of NP if MP = 51
Answer
In triangle ( MNP ) with ( \hat{N} = 90^{\circ} ), we can use the definition of sine in the following way:
- Given ( MP = 51 ), we use the sine relationship: ( \sin M = \frac{NP}{MP} ).
- Rearranging gives us ( NP = MP \cdot \sin M = 51 \cdot \frac{15}{17} = 45.00 ). So, the length of NP is 45.
Step 3
5.2 Simplify to a single term: \( \cos(x - 360^\circ) \sin(90^\circ - x) + \cos^2(z - x) - 1 \)
Answer
- ( \cos(x - 360^\circ) ) simplifies to ( \cos x ) because ( \cos ) is a periodic function.
- ( \sin(90^\circ - x) = \cos x ).
- Therefore, the first term simplifies to ( \cos x \cdot \cos x = \cos^2 x ).
- Thus, the expression can be rewritten as ( \cos^2 x + \cos^2(z - x) - 1 ).
- Finally, applying the Pythagorean identity, the expression simplifies to ( \cos^2(z - x) ).
Step 4
5.3 Consider: \( \sin(2x + 40^\circ) \cos(x + 30^\circ) - \cos(2x + 40^\circ) \sin(2x + 30^\circ) \)
Answer
- We can apply the product-to-sum identities which state that ( \sin A \cos B - \cos A \sin B = \sin(A - B) ).
- In this case, let ( A = 2x + 40^{\circ} ) and ( B = 2x + 30^{\circ} ).
- Therefore, we can simplify to ( \sin((2x + 40^{\circ}) - (2x + 30^{\circ})) = \sin(10^{\circ}) ).
Step 5
Step 6
5.3.2 Determine the general solution of the following equation: \( \sin(2x + 40^{\circ}) \cos(2x + 40^{\circ}) - \cos(2x + 30^{\circ}) = \cos(2x - 20^{\circ}) \)
Answer
- Start with simplifying the equation using product-to-sum identities: ( \frac{1}{2} \sin(2(2x + 40^{\circ})) - \cos(2x + 30^{\circ}) = \cos(2x - 20^{\circ}) ).
- Solve for ( x ) by finding the angles where the equation holds true.
- The general solution can be derived through periodic properties of sine and cosine, leading to: ( x = n\pi + C ), where ( n ) is an integer and ( C ) represents the constants derived from the angles.