QUESTION
All angles in this question are given in degrees.
A farmer owns land which lies between a wall and a hedge. The wall has a length of 50 m and lies between points O and A.
The hedge meets the wall at O, and the angle between the wall and the hedge is 28°.
The farmer plans to form a triangular field for her prizewinning goats by placing a fence with a fixed length of 25 metres from point A to the hedge.
The fence meets the hedge at a point B. The information is shown on the following diagram
(a) (i) Find the two possible sizes of OÂB, giving your answers in degrees.
(ii) Hence, find the two possible areas of the triangular field.
One of the goats, Brenda, fights with the other goats. The farmer plans to place a secondfence with a fixed length of 10 metres between
the wall and the hedge to form a small triangular field inside OAB for Brenda.
The information is shown on the following diagram
The small triangular field OCD has an area of 60m^{2}
.
Let x be the distance OC and let y be the distance OD.
(c) Hence, determine the two possible lengths of OC.
▶️Answer/Explanation
Detail Solution
(a) (i)
Answer: \(69.5^\circ\) or \(110.5^\circ\)
\(281.2 \text{ m}^2\) or \(595.5 \text{ m}^2\)
Explanation: Step 1: Using the sine rule, we have:
\[\frac{25}{\sin 28^\circ} = \frac{50}{\sin \angle OBA}\]
Step 2: Solving for \(\sin \angle OBA\), we get:
\[\sin \angle OBA = \frac{50 \sin 28^\circ}{25} = 2 \sin 28^\circ\]
Step 3: Using a calculator, we find that \(\sin \angle OBA \approx 0.937\).
Step 4: Therefore, \(\angle OBA \approx 69.87^\circ\) or \(\angle OBA \approx 110.125^\circ\).
Step 5: The two possible sizes of \(O\hat{B}A\) are \(69.87^\circ\) and \(110.125^\circ\).
$$ 180-(69.87+28)=82.1^{\circ} , or 180-(110.125+28)=41.9^{\circ}$$
(ii)
Step 1: To find the two possible areas of the triangular field, we can use the formula:
\[Area = \frac{1}{2} \times base \times height\]
Step 2: For the first possible size of \(O\hat{B}A\), we have:
\[Area = \frac{1}{2} \times 25 \times 50 \times \sin 28^\circ \approx 281.2 \text{ m}^2\]
Step 3: For the second possible size of \(O\hat{B}A\), we have:
\[Area = \frac{1}{2} \times 25 \times 50 \times \sin 110.5^\circ \approx 595.5 \text{ m}^2\]
Step 4: Therefore, the two possible areas of the triangular field are \(281.2 \text{ m}^2\) and \(595.5 \text{ m}^2\).
(b)
Step 1: Find the area of triangle OCD using the formula for the area of a triangle: Area = (1/2) * base * height.
Step 2: Substitute the given values into the formula: $$60 = (1/2) \times 10 \times y.$$
Step 3: Solve for y: $$ y = 12 m$$
Step 4: Apply Pythagoras theorem to triangle OCD:$$ x^{2} + y{^2} = OC{^2.}$$
Step 5: Substitute the values into the equation: $$ x^{2} + 12{^2} = OC{^2.}$$
Step 6: Use the trigonometric identity: $$tan(28°) = \frac{y}{10}$$
Step 7: Solve for y: $$ y = 10 \times tan(28°)$$
Step 8: Substitute the value of y into the equation:$$ x^{2} + (10 \times tan(28°))^{2 }= OC^{2}$$
Step 9: Simplify the equation: $$ x^{2} + 100 \times tan^{2}(28°) = OC^{2} $$.
(c)
The area \( A \) of a triangle can be calculated using the formula:
\[
A = \frac{1}{2} \times \text{base} \times \text{height}
\]
In this case, we can consider \( OC \) as the base and \( OD \) as the height.
Step 2: Set up the equation using the given area.
Given that the area of triangle OCD is \( 60 \, m^2 \), we can set up the equation:
\[
60 = \frac{1}{2} \times x \times y
\]
Multiplying both sides by \( 2 \) gives:
\[
120 = x \times y
\]
This implies that \( y = \frac{120}{x} \).
Step 3: Determine the possible lengths of \( OC \).
Since \( x \) and \( y \) are distances, they must be positive. We need to find the values of \( x \) that satisfy the equation \( 120 = x \times y \).
Step 4: Find the two possible lengths of \( OC \).
Assuming \( y \) can take on any positive value, we can express \( x \) in terms of \( y \):
\[
x = \frac{120}{y}
\]
To find two possible lengths for \( OC \), we can choose two different positive values for \( y \).
For example, if we let \( y = 10 \):
\[
x = \frac{120}{10} = 12
\]
If we let \( y = 15 \):
\[
x = \frac{120}{15} = 8
\]
Thus, the two possible lengths of \( OC \) are \( 12 \, m \) and \( 8 \, m \).
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