1.1 To prove that ⃤𝐴𝐵𝐶 is right-angled, we can use the property that a triangle is right-angled if one of its angles is 90 degrees.
First, we can find the slopes of the lines containing the sides of the triangle ⃤𝐴𝐵𝐶:
Slope of 𝐴𝐵 = (2-3)/(5-(-2)) = -1/7
Slope of 𝐵𝐶 = (-1-2)/(2-5) = -1
Slope of 𝐶𝐴 = (3+1)/(-2-2) = -1
From the slopes, we can determine that the angle between 𝐴𝐵 and 𝐶𝐴 is 90 degrees, as the product of their slopes is -1. Therefore, ⃤𝐴𝐵𝐶 is right-angled.
1.2 To calculate the area of ⃤𝐴𝐵𝐶, we can use the formula for the area of a triangle:
Area = 1/2 * base * height.
First, we need to find the base and height of the triangle. The base can be taken as the distance between points 𝐴 and 𝐶, and the height can be taken as the perpendicular distance from point 𝐵 to line 𝐴𝐶.
Base = √[(5-(-2))^2 + (2-3)^2] = √[49 + 1] = √50 = 5√2
Height = Distance from 𝐵 to line 𝐴𝐶 = |(-1-3)/2| = 2
Now, we can plug these values into the formula:
Area = 1/2 * 5√2 * 2 = 5√2.
Therefore, the area of ⃤𝐴𝐵𝐶 is 5√2.
1.3 To calculate the size of θ to the nearest degree, we can use trigonometry.
First, calculate the length of 𝐴𝐶:
Length of 𝐴𝐶 = √[(-2-2)^2 + (3+1)^2] = √[16 + 16] = 4√2.
Now, we can use the cosine function to find θ:
cos θ = adjacent/hypotenuse
cos θ = 5√2 / 4√2
θ = cos^(-1) (5/4) ≈ 36.87 degrees.
Therefore, the size of θ to the nearest degree is approximately 37 degrees.
𝐴(−2;3),𝐵(5;2)and𝐶(2;−1)areverticesof𝐴⃤𝐵𝐶inaCartesianplaneasshown below. θ is the angle of inclination of 𝐴𝐶.
Figure 1: Diagram for question 1.
1.1
Prove that ⃤𝐴𝐵𝐶 is right-angled.
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1.2
Calculatetheareaof𝐴⃤𝐵𝐶.
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1.3
Calculate the size of θ to the nearest degree.
1 answer