NATIONL SENIOR CERTIFICATE: MBTIEMATICS: PAPER I Page 12 of 31 QUESTION 6 (a) Given: $$ \sin 40^{\circ}=\frac{p}{4} $$ Determine the values of the following in terms of p: (1) $$ \sin 50^{\circ}, \cos 10^{\circ}-\cos 50^{\circ} \cdot \sin 10^{\circ} $$ (2) (2) $$ \cos ^{2} 20^{\circ}-\sin ^{2} 20^{\circ} $$ (3) (3) $$ an 50^{\circ} $$ (1) IEA Copright 02024 NATIONL SENICR CERTFOATE: MATHEMATICS: PMPER II Page 13 of 31 (b) In $$ riangle A B C $$ : - $$ A B=10 $$ units. - $$ B C=12 $$ units. - $$ A \hat{B} C=60^{\circ} $$. Determine the length of $$ A C $$.

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NATIONL SENIOR CERTIFICATE: MBTIEMATICS: PAPER I Page 12 of 31 QUESTION 6 (a) Given: $$ \sin 40^{\circ}=\frac{p}{4} $$ Determine the values of the following in terms of p: (1) $$ \sin 50^{\circ}, \cos 10^{\circ}-\cos 50^{\circ} \cdot \sin 10^{\circ} $$ (2) (2) $$ \cos ^{2} 20^{\circ}-\sin ^{2} 20^{\circ} $$ (3) (3) $$ 	an 50^{\circ} $$ (1) IEA Copright 02024 NATIONL SENICR CERTFOATE: MATHEMATICS: PMPER II Page 13 of 31 (b) In $$ 	riangle A B C $$ : - $$ A B=10 $$ units. - $$ B C=12 $$ units. - $$ A \hat{B} C=60^{\circ} $$. Determine the length of $$ A C $$.

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Let's solve the problem step by step.

### Part (a)

We are given that $$ \sin 40^{\circ} = \frac{p}{4} $$. We will use this information to find the required values.

#### (1) Calculate $$ \sin 50^{\circ} $$ and $$ \cos 10^{\circ} - \cos 50^{\circ} \cdot \sin 10^{\circ} $$

Using the sine addition formula:
$$
\sin(50^{\circ}) = \sin(40^{\circ} + 10^{\circ}) = \sin 40^{\circ} \cos 10^{\circ} + \cos 40^{\circ} \sin 10^{\circ}
$$
We know $$ \sin 40^{\circ} = \frac{p}{4} $$. Now we need to find $$ \cos 40^{\circ} $$:
$$
\cos 40^{\circ} = \sqrt{1 - \sin^2 40^{\circ}} = \sqrt{1 - \left(\frac{p}{4}\right)^2} = \sqrt{1 - \frac{p^2}{16}} = \frac{\sqrt{16 - p^2}}{4}
$$

Now, we can express $$ \sin 50^{\circ} $$:
$$
\sin 50^{\circ} = \frac{p}{4} \cos 10^{\circ} + \frac{\sqrt{16 - p^2}}{4} \sin 10^{\circ}
$$

Next, we need to calculate $$ \cos 10^{\circ} - \cos 50^{\circ} \cdot \sin 10^{\circ} $$:
Using the cosine subtraction formula:
$$
\cos(50^{\circ}) = \cos(40^{\circ} + 10^{\circ}) = \cos 40^{\circ} \cos 10^{\circ} - \sin 40^{\circ} \sin 10^{\circ}
$$
Substituting the values:
$$
\cos(50^{\circ}) = \frac{\sqrt{16 - p^2}}{4} \cos 10^{\circ} - \frac{p}{4} \sin 10^{\circ}
$$

Now, we can find $$ \cos 10^{\circ} - \cos 50^{\circ} \cdot \sin 10^{\circ} $$:
$$
\cos 10^{\circ} - \left(\frac{\sqrt{16 - p^2}}{4} \cos 10^{\circ} - \frac{p}{4} \sin 10^{\circ}\right) \sin 10^{\circ}
$$

#### (2) Calculate $$ \cos^2 20^{\circ} - \sin^2 20^{\circ} $$

Using the double angle identity:
$$
\cos^2 20^{\circ} - \sin^2 20^{\circ} = \cos(40^{\circ})
$$
We already found $$ \cos 40^{\circ} = \frac{\sqrt{16 - p^2}}{4} $$.

#### (3) Calculate $$ \tan 50^{\circ} $$

Using the definition of tangent:
$$
\tan 50^{\circ} = \frac{\sin 50^{\circ}}{\cos 50^{\circ}}
$$
We can express $$ \tan 50^{\circ} $$ in terms of $$ p $$ using the values we calculated above.

Now, let's summarize the results for part (a) before moving to part (b).

### Part (b)

In triangle $$ ABC $$:
- $$ AB = 10 $$ units
- $$ BC = 12 $$ units
- $$ \angle ABC = 60^{\circ} $$

We can use the Law of Cosines to find $$ AC $$:
$$
AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle ABC)
$$
Substituting the known values:
$$
AC^2 = 10^2 + 12^2 - 2 \cdot 10 \cdot 12 \cdot \cos(60^{\circ})
$$
Since $$ \cos(60^{\circ}) = \frac{1}{2} $$:
$$
AC^2 = 100 + 144 - 2 \cdot 10 \cdot 12 \cdot \frac{1}{2}
$$
$$
AC^2 = 100 + 144 - 120
$$
$$
AC^2 = 124
$$
$$
AC = \sqrt{124} = 2\sqrt{31}
$$

Now, let's summarize the results for part (b).

### Final Summary

**Part (a):**
1. $$ \sin 50^{\circ} = \frac{p}{4} \cos 10^{\circ} + \frac{\sqrt{16 - p^2}}{4} \sin 10^{\circ} $$
2. $$ \cos^2 20^{\circ} - \sin^2 20^{\circ} = \frac{\sqrt{16 - p^2}}{4} $$
3. $$ \tan 50^{\circ} = \frac{\sin 50^{\circ}}{\cos 50^{\circ}} $$

**Part (b):**
- The length of $$ AC = 2\sqrt{31} $$ units.
**Part (a):** 1. $$ \sin 50^{\circ} = \frac{p}{4} \cos 10^{\circ} + \frac{\sqrt{16 - p^2}}{4} \sin 10^{\circ} $$ 2. $$ \cos^2 20^{\circ} - \sin^2 20^{\circ} = \frac{\sqrt{16 - p^2}}{4} $$ 3. $$ \tan 50^{\circ} = \frac{\sin 50^{\circ}}{\cos 50^{\circ}} $$ **Part (b):** - The length of $$ AC = 2\sqrt{31} $$ units.

NATIONL SENIOR CERTIFICATE: MBTIEMATICS: PAPER I
Page 12 of 31

QUESTION 6
(a) Given:

Determine the values of the following in terms of p:
(1)
(2)
(2)
(3)
(3)
(1)
IEA Copright 02024
NATIONL SENICR CERTFOATE: MATHEMATICS: PMPER II
Page 13 of 31
(b) In :

units.

units.

.

Determine the length of .

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NATIONL SENIOR CERTIFICATE: MBTIEMATICS: PAPER I Page 12 of 31 QUESTION 6 (a) Given: Determine the values of the following in terms of p: (1) (2) (2) (3) (3) (1) IEA Copright 02024 NATIONL SENICR CERTFOATE: MATHEMATICS: PMPER II Page 13 of 31 (b) In : units. units. . Determine the length of .