a triangle ABC, BC = 8cm, AC = 12cm and angle ABC = 120º.
Calculate the of AB, correct to one decimal place.
(b) If BC is the base of the triangle, calculate, correct to one decimal place:
(i) the perpendicular height of the triangle;
(ii) the area of the triangle;
(iii) the size angle ACB
AB = 12 cos60º
(i) 12 sin60º
(ii) A = 1/2 * BC * (i)
(iii) sinC = (ii)/((i)+8)
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To find the length of side AB of the triangle ABC, we can use the law of cosines. The law of cosines states:
c^2 = a^2 + b^2 - 2ab*cos(C)
Where c is the length of side c (in this case, side AB), a and b are the lengths of the other two sides (in this case, sides BC and AC), and C is the angle opposite side c.
Plugging in the given values, we have:
AB^2 = 8^2 + 12^2 - 2*8*12*cos(120º)
Simplifying:
AB^2 = 64 + 144 - 192*cos(120º)
AB^2 = 208 - 192*(-0.5)
AB^2 = 208 + 96
AB^2 = 304
Taking the square root of both sides:
AB ≈ √(304) ≈ 17.5 cm (rounded to one decimal place)
Therefore, the length of AB is approximately 17.5 cm.
(b) Now let's calculate the other quantities:
(i) To find the perpendicular height of the triangle (h), we can use the formula:
h = b*sin(C)
In this case, b is the length of the base (BC), and C is the angle opposite the height (angle ABC). Plugging in the values:
h = 8*sin(120º)
h = 8*(-0.866)
h ≈ -6.928 cm (rounded to one decimal place)
Since height cannot be negative, the perpendicular height is approximately 6.9 cm. This means that the height is 6.9 cm above point A, in a direction perpendicular to base BC.
(ii) The area of a triangle can be calculated using the formula:
Area = (1/2)*base*height
Substituting the known values:
Area = (1/2)*8*6.9
Area = 27.6 cm^2 (rounded to one decimal place)
Therefore, the area of the triangle is approximately 27.6 cm^2.
(iii) To find the size of angle ACB, we can use the law of cosines again. Rearranging the previous equation:
cos(C) = (a^2 + b^2 - c^2) / (2ab)
Plugging in the given values:
cos(ACB) = (8^2 + 12^2 - AB^2) / (2*8*12)
cos(ACB) = (64 + 144 - 304) / (192)
cos(ACB) = -96 / 192
cos(ACB) = -0.5
To find the angle, we can use the inverse cosine function (cos^-1):
ACB ≈ cos^-1(-0.5)
ACB ≈ 120º
Therefore, the size of angle ACB is 120º.