M21-5-MATHY-HP2-ENG-TZ2-XX

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Keycap Digit Two

Paper 2

Question 1 - MaximumMarks - 15

A farmer owns a field in the shape of a triangle ABC. The lengths of the sides of this triangle are given: AB = 650 meters, AC = 1005 meters, and BC = 1225 meters.

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In simpler terms, this means that we have a triangle with three sides. The distance between points A and B is 650 meters, between points A and C is 1005 meters, and between points B and C is 1225 meters.

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Question 1 - Image

Question 1 - A

Determine the measure of angle&nbsp;\( (\angle ACB)\).

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This question is asking for the measurement of the angle formed at point (C) between points (A) and (B). You need to find the size of this angle in degrees or radians, depending on the context provided.

Question 1 - A - Video - Coming soon

Question 1 - A - Answer

To determine the size of angle AĈB in triangle ABC, we apply the Cosine Rule, which relates the lengths of a triangle&#39;s sides to the cosine of one of its angles. Given AB = 650m, AC = 1005m, and BC = 1225m, the Cosine Rule formula for AĈB is:

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\([ \cos(AĈB) = \frac{1005^2 + 1225^2 - 650^2}{2 \times 1005 \times 1225} ]\)

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Substituting the given values and calculating, we obtain:

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\([ AĈB = \cos^{-1}\left(\frac{1005^2 + 1225^2 - 650^2}{2 \times 1005 \times 1225}\right) ]\)

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This calculation yields \((AĈB = 32.0^\circ)\) or (AĈB = 0.558) radians.

Question 1 - A - Answer - EC

The candidate correctly applied the Cosine Rule to find the size of angle AĈB. The formula was accurately used with the given side lengths, and the mathematical operations were carried out precisely. This approach demonstrates a solid understanding of the relationships within a triangle and the application of trigonometric principles. The answer provided aligns exactly with the expected result as per the marking scheme, which calls for either a degree measure of (32.0&deg;) or a radian measure of (0.558), showcasing the candidate&#39;s ability to correctly perform and interpret trigonometric calculations. The work is presented in a clear, logical manner, with each step of the calculation properly detailed, fulfilling the criteria for full marks.

Question 1 - B

A town is planning to extend a highway, creating a segment that will pass through a field, intersecting at points D and E. It&#39;s known that the length from D to C is 210 meters and that the angle CED measures 100 degrees.

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The objective is to determine the distance between D and E.

IB Math AI HL - May 21 TZ2 - Paper 2 - question  - 1 - B.svg

Question 1 - B - Image

Question 1 - B - Video - Coming soon

Question 1 - B - Answer

The local town is planning to build a highway that will intersect the borders of the field at points D and E, where (DC = 210, m) and \(( \angle CED = 100^\circ )\), as shown in the diagram below.

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First, we need to determine the length of (DE). Using the given diagram, we can apply the law of cosines in triangle \(( \triangle ACB )\):

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\([ \angle ACB = \cos^{-1} \left( \frac{1005^2 + 1225^2 - 650^2}{2 \times 1005 \times 1225} \right) ]\)

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Calculating the value inside the inverse cosine function:

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\([\angle ACB = \cos^{-1} \left( \frac{1005^2 + 1225^2 - 650^2}{2 \times 1005 \times 1225} \right) \approx \cos^{-1} (0.558) \approx 32^\circ \)&nbsp;(rounded to two decimal places).

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Next, we use the sine rule in triangle \(( \triangle CDE )\):

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\([\frac{DE}{\sin 31.998^\circ} = \frac{210 \, m}{\sin 100^\circ}]\)

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Simplifying the equation to find (DE):

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\([ DE = \frac{210 \times \sin 31.998^\circ}{\sin 100^\circ} \approx \frac{210 \times 0.5299}{0.9848} \approx 112.99 \, m]\) (rounded to two decimal places).

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Therefore, the length of (DE) is approximately (113 m).

Question 1 - B - Answer - EC

The candidate correctly used the law of cosines to determine \(( \angle ACB)\) in the given triangle. The calculation inside the inverse cosine function was shown step-by-step, leading to the angle measurement of \(( 32^\circ )\), which is accurately rounded to two decimal places. This demonstrates a clear understanding of trigonometric principles and careful computation.

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In the subsequent steps, the candidate applied the sine rule in triangle \(( \triangle CDE )\). They correctly set up the ratio and rearranged the equation to solve for (DE). The intermediate steps were detailed, showcasing the conversion from angles to their sine values, and the final multiplication was executed correctly.

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The final answer, \(( DE \approx 113 , m)\), is within the acceptable range and rounded appropriately. This level of precision and attention to detail is commendable and indicative of a high standard of mathematical problem-solving skills.

Question 1 - C

The local authorities are interested in transforming a triangular section of a farmer&#39;s land into a public parking space. This triangular section is denoted by DCE. In a fair exchange, the farmer will receive a new tract of land, triangle ADF, which has an equivalent size to triangle DCE, with the stipulation that point F is directly in line with points D and E.

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The challenge lies in calculating the magnitude of the farmer&#39;s land that is up for exchange, essentially the area of triangle DCE.

Question 1 - C - Video - Coming soon

Question 1 - C - Answer

To find the area of triangle (DCE), we first need to determine the measures of all relevant angles and then use the appropriate area formula for a triangle given its sides and angles.

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We are given that (DC = 210 , m) and \(( \angle CED = 100^\circ)\). We can find \(( \angle DCE)\)&nbsp;as follows:

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\(( \angle DCE = 180^\circ - (100^\circ + \angle EDC))\)

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Since \(( \angle EDC = \alpha)\), where \(( \alpha)\) is a part of the given solution, and we know that \(( \angle DCE )\) is calculated as:

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\(( \angle DCE = 180^\circ - (100^\circ + \alpha) = 48.0019^\circ )\)&nbsp;or \(( 48^\circ \, 0' \, 11.4'' )\)

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Next, we use the formula for the area of a triangle when two sides and the included angle are known:

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\(( \text{Area} = \frac{1}{2} \times a \times b \times \sin(\theta))\)

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Here, (a = 112.9937 ,m) (DE) and ( b = 210 , m) (DC), and \(( \theta = 48.002^\circ)\). Substituting these values into the formula:

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\(( \text{Area} = \frac{1}{2} \times 112.9937 \times 210 \times \sin(48.002^\circ))\)

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Calculating this, we find:

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(Area = 8820 , m2) or approximately (8817.18 ,m2)

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Therefore, the area of triangle (DCE) is 8820 square meters.

Question 1 - C - Answer - EC

The candidate has correctly identified and used the given sides and angles of triangle (DCE) to apply the appropriate formula for the area of a triangle. All intermediate steps are shown, including the calculation of \(( \angle DCE)\) and the application of the sine function in the area formula.

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The calculations are accurate and the final answer is clearly stated. This answer demonstrates a solid understanding of trigonometric principles and their application in solving geometric problems. As a result, the candidate&#39;s solution deserves full marks.

Question 1 - D

Estimate the distance function (DF) considering that the highway has a width of zero.

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This means you should calculate the distance function assuming the highway is a one-dimensional line.

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Question 1 - D - Video - Coming soon

Question 1 - D - Answer

The area of triangle DCE can be found using the formula for the area of a triangle:

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\((\frac{1}{2} \times \text{base} \times \text{height} \times \sin(\text{angle}))\).

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In this case, the base is DC = 210m, the height is EC = 795m (since EC = AC - DC = 1005m - 210m), and the angle C&Ecirc;D is 100&deg;.

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Thus, the area of triangle DCE is

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\((\frac{1}{2} \times 210 \times 795 \times \sin(100°))\).

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This value can be calculated to be

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approximately 8817.18 square meters.

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To find the length of DF, we equate this area to the area of triangle ADF, which has the same height (EC = 795m) but an unknown base DF. Therefore, the area of triangle ADF is

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\((\frac{1}{2} \times DF \times 795 \times \sin(48.002°))\).

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Equating this to the area of triangle DCE, we have:

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\((\frac{1}{2} \times DF \times 795 \times \sin(48.002°) = 8817.18)\)

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Solving for DF, we find that DF &asymp; 29.8m.

Question 1 - D - Answer - EC

The candidate correctly applied the formula for the area of a triangle, using the known sides and angle. The use of \((\frac{1}{2} \times \text{base} \times \text{height} \times \sin(\text{angle}))\) to calculate the area of triangle DCE was accurate, taking into account the geometry of the given triangle.

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In the next step, the candidate adeptly equated the area of triangle DCE to the area of triangle ADF. This step is crucial as it ensures that the areas of the two triangles are the same, which is a key requirement in the question. The use of sine law for the second triangle, considering the given angle C&Ecirc;D and adjusting it to 48.002&deg; for the triangle ADF, was a precise and appropriate method.

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The arithmetic calculation for DF was executed correctly, leading to a final answer of DF &asymp; 29.8m. This solution is in agreement with the marking scheme, thus warranting full marks. The candidate&rsquo;s methodical approach and accurate calculations are commendable and demonstrate a thorough understanding of the principles of trigonometry as applied to real-world problems.

Question 2 - MaximumMarks - 16

When looking at male Persian cats, it&#39;s noted that their body weights tend to follow a consistent pattern. On average, these cats have a weight of 6.1 kilograms.

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There&#39;s also a measure of how much the weights can vary from this average, known as the variance, which in this case is 0.5 kilograms squared. This tells us about the natural weight differences you&#39;d expect to see among these cats.

IB MATH AI HL, Paper 2, May, 2021, TZ2, Solved Past Paper