In the diagram $ riangle EGF$ is such that $ riangle EGF = 129^ heta$, |EG| = 8$ and |FG| = 10. Calculate the area of the triangle $EFG$, giving your answer correct to one decimal place. In the diagram $ABC$ is a right-angled triangle, with $AC$ perpendicular to $BC$. |AC| = $rac{2 oot{2}}{2}$ and |BC| = $rac{3}{ oot{3}}$. Calculate (i) |AB|, leaving your answer in surd form (ii) |$ heta ABC|$, correct to the nearest degree.

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In the diagram $ riangle EGF$ is such that $ riangle EGF = 129^ heta$, |EG| = 8$ and |FG| = 10 - Junior Cycle Mathematics - Question 5 - 2012

Question 5

In the diagram $ riangle EGF$ is such that $ riangle EGF = 129^ heta$, |EG| = 8$ and |FG| = 10. Calculate the area of the triangle $EFG$, giving your answer correc... show full transcript

Worked Solution & Example Answer:In the diagram $ riangle EGF$ is such that $ riangle EGF = 129^ heta$, |EG| = 8$ and |FG| = 10 - Junior Cycle Mathematics - Question 5 - 2012

Step 1

Calculate the area of the triangle EFG

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Answer

To find the area of triangle $EFG$ , we use the formula:

$A = \frac{1}{2} \times b \times h \times \sin(C)$

Where:

$b = |EG| = 8$ ,
$h = |FG| = 10$ ,
$C = 129^ heta$

Substituting the values:

$A = \frac{1}{2} \times 8 \times 10 \times \sin(129^ heta)$

First, we find $ext{sin}(129^ heta)$ . Using a calculator, we find:

Now substituting this into the area formula: $$A = \frac{1}{2} \times 8 \times 10 \times 0.847 \approx 33.88$$ Thus, the area is approximately: $$A \approx 33.9 \, ext{square units (to one decimal place)}$$

Step 2

Calculate (i) |AB|, leaving your answer in surd form

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Answer

Given that angle $ABC$ is a right angle, we can apply the Pythagorean theorem:

$|AB|^2 = |AC|^2 + |BC|^2$

Substituting the lengths:

$|AB|^2 = \left(\frac{2\sqrt{2}}{2}\right)^2 + \left(\frac{3}{\sqrt{3}}\right)^2$

Calculating each term:

$|AC|^2 = \frac{2^2 \cdot 2}{4} = 2$
$|BC|^2 = \frac{3^2}{3} = 3$

Thus,

$|AB|^2 = 2 + 3 = 5$

Taking the square root,

$|AB| = \sqrt{5}$

Step 3

Calculate (ii) |$ heta ABC|$, correct to the nearest degree

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Answer

Using the sine rule for $riangle ABC$ , we know:

$\tan\theta = \frac{opposite}{adjacent} = \frac{|AC|}{|BC|}$

Substituting the values:

$\tan\theta = \frac{\frac{2\sqrt{2}}{2}}{\frac{3}{\sqrt{3}}}$

First simplifying:

$\tan\theta = \frac{\sqrt{2}}{\frac{3}{\sqrt{3}}} = \frac{\sqrt{2} \cdot \sqrt{3}}{3} = \frac{\sqrt{6}}{3}$

Now, taking the inverse tangent:

$\theta = \tan^{-1}\left(\frac{\sqrt{6}}{3}\right) \approx 28.6^ heta$

Rounding to the nearest degree gives:

$\theta \approx 29^ heta$

In the diagram $ riangle EGF$ is such that $ riangle EGF = 129^ heta$, |EG| = 8$ and |FG| = 10 - Junior Cycle Mathematics - Question 5 - 2012

Worked Solution & Example Answer:In the diagram $ riangle EGF$ is such that $ riangle EGF = 129^ heta$, |EG| = 8$ and |FG| = 10 - Junior Cycle Mathematics - Question 5 - 2012

Calculate the area of the triangle EFG

Calculate (i) |AB|, leaving your answer in surd form

Calculate (ii) |$ heta ABC|$, correct to the nearest degree

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