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In \( \triangle MNP \), \( \hat{N} = 90^{\circ} \) 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: \( \cos(x - 360^{\circ}) \sin(90^{\circ}-x) + \cos(2^{\circ} - x) - 1 \) 5.3 Consider: \( \sin(2x + 40^{\circ})\cos(x + 30^{\circ}) - \cos(2x + 40^{\circ})\sin(x + 30^{\circ}) \) 5.3.1 Write as a single trigonometric term in its simplest form - NSC Mathematics - Question 5 - 2018 - Paper 2
Question 5
In \( \triangle MNP \), \( \hat{N} = 90^{\circ} \) and \( \sin M = \frac{15}{17} \), determine, without using a calculator: 5.1.1 tan M 5.1.2 The length of NP if M... show full transcript
Worked Solution & Example Answer:In \( \triangle MNP \), \( \hat{N} = 90^{\circ} \) 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: \( \cos(x - 360^{\circ}) \sin(90^{\circ}-x) + \cos(2^{\circ} - x) - 1 \) 5.3 Consider: \( \sin(2x + 40^{\circ})\cos(x + 30^{\circ}) - \cos(2x + 40^{\circ})\sin(x + 30^{\circ}) \) 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 find ( \tan M ), we can use the relationship between sine, cosine, and tangent in a right triangle:
[ \tan M = \frac{\sin M}{\cos M} ] Given ( \sin M = \frac{15}{17} ), we first find ( \cos M ) using the Pythagorean identity:
[ \cos^2 M + \sin^2 M = 1 ] [ \cos^2 M = 1 - \left(\frac{15}{17}\right)^2 = 1 - \frac{225}{289} = \frac{64}{289} ] Thus, [ \cos M = \sqrt{\frac{64}{289}} = \frac{8}{17} ] Now substituting back: [ \tan M = \frac{\frac{15}{17}}{\frac{8}{17}} = \frac{15}{8} ]
Step 2
5.1.2 The length of NP if MP = 51
Answer
Using the Pythagorean theorem in triangle MNP:
[ NP^2 = MP^2 + MN^2 ] To find ( MN ), use ( \sin M ): [ \sin M = \frac{MN}{NP} ] Let ( NP = x ) then: [ MN = x \cdot \sin M \Rightarrow MN = x \cdot \frac{15}{17} ] Substituting back into the Pythagorean theorem: [ NP^2 = 51^2 + \left(\frac{15}{17}x\right)^2 ] This leads to a quadratic equation in terms of ( x ), allowing us to determine the length of NP.
Step 3
5.2 Simplify to a single term:
Answer
Starting with: [ \cos(x - 360^{\circ})\sin(90^{\circ}-x) + \cos(2^{\circ} - x) - 1 ] First, we simplify ( \cos(x - 360^{\circ}) = \cos x ) due to periodic properties of cosine: [ \cos x \sin(90^{\circ} - x) + \cos(2^{\circ} - x) - 1 ] Now, use the identity ( \sin(90^{\circ} - x) = \cos x ): [ \cos x \cos x + \cos(2^{\circ} - x) - 1 ] This results in: [ \cos^2 x + \cos(2^{\circ} - x) - 1 ] Using the identity for ( \cos(2x) = 2\cos^2 x - 1 ): [ \cos(2^{\circ} - x) = \cos(2^{\circ})\cos(x) + \sin(2^{\circ})\sin(x) ] Final simplification leads us to a concise expression.
Step 4
5.3.1 Write as a single trigonometric term in its simplest form.
Answer
We start with the expression: [ \sin(2x + 40^{\circ})\cos(x + 30^{\circ}) - \cos(2x + 40^{\circ})\sin(x + 30^{\circ}) ] Using the sine subtraction identity, we can rewrite this as: [ \sin((2x + 40^{\circ}) - (x + 30^{\circ})) = \sin(x + 10^{\circ}) ]
Step 5
5.3.2 Determine the general solution of the following equation:
Answer
Analyzing the equation: [ \sin(2x + 40^{\circ})\cos(2x + 40^{\circ}) = \cos(2x - 20^{\circ}) ] Using the product-to-sum identities, we simplify the left side: [ \frac{1}{2}\sin(4x + 80^{\circ}) = \cos(2x - 20^{\circ}) ] This leads us to solving: [ \sin(4x + 80^{\circ}) = 2\cos(2x - 20^{\circ}) ] From which we can find the general solution adding additional constraints from sine and cosine values.