In a triangle $ABC$, the lengths of the sides are $a$, $b$ and $c$. Using a formula for the area of a triangle, or otherwise, prove that $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.$$ In a triangle $XYZ$, $|XY| = 5$ cm, $|XZ| = 3$ cm and $|\angle XYZ| = 27^{\circ}$. (i) Find the two possible values of $|\angle XZY|$. Give your answers correct to the nearest degree. (ii) Draw a sketch of the triangle $XYZ$, showing the two possible positions of the point $Z$. In the case that $|\angle XZY| < 90^{\circ}$, write down $|\angle ZXY|$, and hence find the area of the triangle $XYZ$, correct to the nearest integer.

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In a triangle $ABC$, the lengths of the sides are $a$, $b$ and $c$ - Leaving Cert Mathematics - Question 5 - 2013

Question 5

In a triangle $ABC$, the lengths of the sides are $a$, $b$ and $c$. Using a formula for the area of a triangle, or otherwise, prove that $$\frac{a}{\sin A} = \frac... show full transcript

Worked Solution & Example Answer:In a triangle $ABC$, the lengths of the sides are $a$, $b$ and $c$ - Leaving Cert Mathematics - Question 5 - 2013

Step 1

Find the two possible values of $|\angle XZY|$

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Answer

To find the two possible values of $|\angle XZY|$ , we can apply the sine rule, which states that:

$\frac{|XY|}{\sin |XZ|} = \frac{|XZ|}{\sin |XY|}$

Given:

$|XY| = 5 \text{ cm}$
$|XZ| = 3 \text{ cm}$
$|\angle XYZ| = 27^{\circ}$

Using the formula: $\frac{5}{\sin 27^{\circ}} = \frac{3}{\sin |\angle XZY|} \implies 3 \sin 27^{\circ} = 5 \sin |\angle XZY|$

Solving for $|\angle XZY|$ : $\sin |\angle XZY| = \frac{3 \sin 27^{\circ}}{5}$ Calculating: $\sin |\angle XZY| = \frac{3 \times 0.454}{5} = 0.2724$

Thus, $|\angle XZY| = \sin^{-1}(0.2724) \approx 16^{\circ} \quad \text{(first solution)}$ And for the second value: $|\angle XZY| = 180^{\circ} - 16^{\circ} \approx 164^{\circ}$

Therefore, the two possible values of $|\angle XZY|$ are approximately $16^{\circ}$ and $164^{\circ}$ .

Step 2

Draw a sketch of the triangle $XYZ$, showing the two possible positions of the point $Z$

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Answer

The sketch will depict triangle $XYZ$ with positions $Y$ , $X$ , and two possible locations for point $Z$ , illustrating how $|\angle XZY|$ could be either acute or obtuse based on the calculated angles.

For $|\angle XZY| = 16^{\circ}$ : The angle at $Z$ is sharp, making $Z$ close to line $XY$ .
For $|\angle XZY| = 164^{\circ}$ : The angle at $Z$ is obtuse, positioning $Z$ on the opposite side of line segment $XY$ .

Step 3

In the case that $|\angle XZY| < 90^{\circ}$, write down $|\angle ZXY|$, and hence find the area of the triangle $XYZ$, correct to the nearest integer.

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Answer

If $|\angle XZY| < 90^{\circ}$ , we use the property: $|\angle ZXY| = 180^{\circ} - (|\angle XYZ| + |\angle XZY|)$ Substituting the known angles: $|\angle ZXY| = 180^{\circ} - (27^{\circ} + 16^{\circ}) = 137^{\circ}$

To find the area of triangle $XYZ$ , we use the formula: $\Delta = \frac{1}{2} \cdot |XY| \cdot |XZ| \cdot \sin |\angle XYZ|.$ Given:

$|XY| = 5$ cm
$|XZ| = 3$ cm
$|\angle XYZ| = 27^{\circ}$ ,

Calculating: $\Delta = \frac{1}{2} \cdot 5 \cdot 3 \cdot \sin(27^{\circ}) \approx 0.5 \cdot 5 \cdot 3 \cdot 0.454 \approx 3.415\text{ cm}^2$ Rounding to the nearest integer: The area of triangle $XYZ$ is approximately $3$ cm².

In a triangle $ABC$, the lengths of the sides are $a$, $b$ and $c$ - Leaving Cert Mathematics - Question 5 - 2013

Worked Solution & Example Answer:In a triangle $ABC$, the lengths of the sides are $a$, $b$ and $c$ - Leaving Cert Mathematics - Question 5 - 2013

Find the two possible values of $|\angle XZY|$

Draw a sketch of the triangle $XYZ$, showing the two possible positions of the point $Z$

In the case that $|\angle XZY| < 90^{\circ}$, write down $|\angle ZXY|$, and hence find the area of the triangle $XYZ$, correct to the nearest integer.

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