The diagram below shows a triangular patch of ground $ \triangle GHS $, with $|SH| = 58 $ m, $|GH| = 30 $ m and $|GHS| = 68^\circ $. The circle is a helicopter pad. It is the incircle of $ \triangle GHS $ and has centre $ P $. (a) Find $|SG|$. Give your answer in metres, correct to 1 decimal place. (b) Find $ \angle HSG $. Give your answer in degrees, correct to 2 decimal places. (c) Find the area of $ \triangle GHS $. Give your answer in $ m^2 $, correct to 2 decimal places. (d) (i) Find the area of $ \triangle AHSP $, in terms of $ r $, where $ r $ is the radius of the helicopter pad. (ii) Show that the area of $ \triangle GHS $, in terms of $ r $, can be written as $ 71-2r \,m^2 $. (iii) Find the value of $ r $. Give your answer in metres, correct to 1 decimal place. (e) $[S7]$ is a vertical pole at the point $ S $. The angle of elevation of the top of the pole from the point $ P $ is 14\degree. Find the height of the pole. Give your answer, in metres, correct to 1 decimal place.

Leaving Cert Mathematics Trigonometry: The diagram below shows a triang

The diagram below shows a triangular patch of ground $ \triangle GHS $, with $|SH| = 58 $ m, $|GH| = 30 $ m and $|GHS| = 68^\circ $ - Leaving Cert Mathematics - Question 9 - 2019

Question 9

$The-diagram-below-shows-a-triangular-patch-of-ground-$-\triangle-GHS-$,-with-$|SH|-=-58-$-m,-$|GH|-=-30-$-m-and-$|GHS|-=-68^\circ-$-Leaving Cert Mathematics-Question 9-2019.png$

The diagram below shows a triangular patch of ground $ \triangle GHS $, with $|SH| = 58 $ m, $|GH| = 30 $ m and $|GHS| = 68^\circ $. The circle is a helicopt... show full transcript

Worked Solution & Example Answer:The diagram below shows a triangular patch of ground $ \triangle GHS $, with $|SH| = 58 $ m, $|GH| = 30 $ m and $|GHS| = 68^\circ $ - Leaving Cert Mathematics - Question 9 - 2019

Step 1

Find $|SG|$. Give your answer in metres, correct to 1 decimal place.

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Answer

To find (|SG|), we can use the cosine rule:

$|SG|^2 = |SH|^2 + |GH|^2 - 2|SH| |GH| \cos(\angle GHS)$

Substituting the known values: $|SG|^2 = 30^2 + 58^2 - 2 \cdot 30 \cdot 58 \cdot \cos(68^\circ)$

Calculating each part:

(30^2 = 900)
(58^2 = 3364)
(\cos(68^\circ) \approx 0.3746) leads to (-2 \cdot 30 \cdot 58 \cdot 0.3746 \approx -1293.6)

Now, summing these results: $|SG|^2 = 900 + 3364 - 1293.6\ = 2960.4\Rightarrow |SG| \approx 54.4 \text{ m}$

Step 2

Find $ \angle HSG $. Give your answer in degrees, correct to 2 decimal places.

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Answer

Using the sine rule:

$\frac{|GH|}{\sin(\angle HSG)} = \frac{|SG|}{\sin(\angle GHS)}$

Substituting our known values gives:

$\frac{30}{\sin(\angle HSG)} = \frac{54.4}{\sin(68^\circ)}$

From which solving for ( \angle HSG ) provides:

$\sin(\angle HSG) \approx \frac{30 \cdot \sin(68^\circ)}{54.4} \approx 0.51131$

Thus, ( \angle HSG \approx 30.75^\circ ).

Step 3

Find the area of $ \triangle GHS $. Give your answer in $ m^2 $, correct to 2 decimal places.

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Answer

The area ( A ) can be found using the formula:

$A = \frac{1}{2} |SH| \cdot |GH| \cdot \sin(\angle GHS)$

Substituting our values:

$A = \frac{1}{2} \cdot 58 \cdot 30 \cdot \sin(68^\circ)$

Calculating yields: $A \approx 806.65 \text{ m}^2 \Rightarrow 806.65 \approx 806.65 \text{ m}^2$

Step 4

Find the area of $ \triangle AHSP $, in terms of $ r $, where $ r $ is the radius of the helicopter pad.

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Answer

The area can be defined as:

$A_{AHSP} = \frac{1}{2} \cdot |SH| \cdot r$

Thus, we denote: $A_{AHSP} = \frac{1}{2} (58)(r) = 29r$ .

Step 5

Show that the area of $ \triangle GHS $, in terms of $ r $, can be written as $ 71-2r \,m^2 $.

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Answer

The area can be computed using the known triangles. The total area becomes:

$A_{GHS} = \frac{1}{2} \cdot 30 \cdot (54.4-2r) + A_{AHSP}$

Simplifying: $\Rightarrow 15r + 27r + 29r = 71 - 2r.$

Step 6

Find the value of $ r $. Give your answer in metres, correct to 1 decimal place.

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Answer

Setting equations from previous area calculations yields: $71 - 2r = 806 - 62\Rightarrow 806-62 = 71 + 2r$

Solving this provides: $r = \frac{806 - 62}{71} \approx 11.3 \text{ m}$ .

Step 7

Find the height of the pole.

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Answer

Using the tangent ratio:

$\tan(14^\circ) = \frac{|ST|}{|PS|}\Rightarrow |ST| = |PS|\cdot \tan(14^\circ).$

Substituting evaluates: $|PS| = \frac{11.3 \cdot \tan(14^\circ)}{1}\Rightarrow |PS| \approx 42.51 \text{ m}$ .

The diagram below shows a triangular patch of ground \( \triangle GHS \), with \(|SH| = 58 \) m, \(|GH| = 30 \) m and \(|GHS| = 68^\circ \) - Leaving Cert Mathematics - Question 9 - 2019

Worked Solution & Example Answer:The diagram below shows a triangular patch of ground \( \triangle GHS \), with \(|SH| = 58 \) m, \(|GH| = 30 \) m and \(|GHS| = 68^\circ \) - Leaving Cert Mathematics - Question 9 - 2019

Find \(|SG|\). Give your answer in metres, correct to 1 decimal place.

Find \( \angle HSG \). Give your answer in degrees, correct to 2 decimal places.

Find the area of \( \triangle GHS \). Give your answer in \( m^2 \), correct to 2 decimal places.

Find the area of \( \triangle AHSP \), in terms of \( r \), where \( r \) is the radius of the helicopter pad.

Show that the area of \( \triangle GHS \), in terms of \( r \), can be written as \( 71-2r \,m^2 \).

Find the value of \( r \). Give your answer in metres, correct to 1 decimal place.

Find the height of the pole.

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