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Find the distance x in the diagram below (not to scale) - Leaving Cert Mathematics - Question 6 - 2017

Question 6

Find the distance x in the diagram below (not to scale). Give your answer correct to 2 decimal places. Find the distance y in the diagram below (not to scale). Give... show full transcript

Worked Solution & Example Answer:Find the distance x in the diagram below (not to scale) - Leaving Cert Mathematics - Question 6 - 2017

Step 1

Find the distance x in the diagram below (not to scale).

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Answer

To find the distance $x$ , we can use the sine rule. According to the sine rule:

$\frac{x}{\sin(63^\circ)} = \frac{10}{\sin(65^\circ)}$

Rearranging this formula:

$x = \frac{10 \cdot \sin(63^\circ)}{\sin(65^\circ)}$

Now, using a calculator, we get:

Calculate $\sin(63^\circ) \approx 0.8910$ .
Calculate $\sin(65^\circ) \approx 0.9063$ .
Substitute into the equation:

$x = \frac{10 \cdot 0.8910}{0.9063} \approx 8.84 \text{ cm}$

Thus, the distance x is approximately 8.84 cm.

Step 2

Find the distance y in the diagram below (not to scale).

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Answer

To calculate the distance $y$ , we can again apply the cosine rule:

$y^2 = a^2 + b^2 - 2ab \cdot \cos(C)$

Where:

$a = 8.5$ cm
$b = 10.2$ cm
$C = 53.8^\circ$

Thus:

$y^2 = (8.5)^2 + (10.2)^2 - 2 \cdot (8.5) \cdot (10.2) \cdot \cos(53.8^\circ)$

Calculating each term:

$y^2 = 72.25 + 104.04 - 2 \cdot 8.5 \cdot 10.2 \cdot 0.5736$ .
Then, plug in the values: $y^2 = 72.25 + 104.04 - 84.04 \approx 92.25$
Finally, take the square root: $y \approx \sqrt{92.25} \approx 9.61 \text{ cm}$

Therefore, the distance y is approximately 9.61 cm.

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Find the distance x in the diagram below (not to scale) - Leaving Cert Mathematics - Question 6 - 2017

Worked Solution & Example Answer:Find the distance x in the diagram below (not to scale) - Leaving Cert Mathematics - Question 6 - 2017

Find the distance x in the diagram below (not to scale).

Find the distance y in the diagram below (not to scale).

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