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Problem #3

The diagram shows an obtuse triangle
ABC with AB = 57 cm, AC = 48 cm and
BÂC=117 degrees
Calculate:
a the length BC
(3 marks)
b the area of triangle ABC (2 marks)
c the size of the angle ABC. (3 marks)

Problem #3 The diagram shows an obtuse triangle ABC with AB = 57 cm, AC = 48 cm and-example-1
User Gab
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2 Answers

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User Henry Blyth
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a) The Length of BC ≈ 99.72 cm

b) The area of Triangle ABC ≈ 1482.135 cm²

c) The size of the angle ABC ≈ 72.86°

a) Length of BC using Law of Cosines:

The Law of Cosines states that for any triangle ABC with sides a, b, and c, and angle C opposite side c:

c² = a² + b² - 2ab cos(C)

In this case, we have:

BC² = AB² + AC² - 2ABAC cos(BAC)

Substitute the given values:

BC² = 57² + 48² - 2 * 57 * 48 * cos(117°)

Calculate BC:

BC =
√(57^2 + 48^2 * 57 * 48 * cos117)

BC ≈
√(3249 + 2304 + 4392)

BC ≈
√(9945)

BC ≈ 99.72 cm

b) Area of Triangle ABC:

The area (A) of a triangle can be calculated using the formula:

A = 1/2 * base * height

AB as the base, the length of AB = 57 cm, and find the height.

We can use the sine of angle BAC to find the height:

height = AB * sin(BAC) = 57 * sin(117°)

The area:

Area = 1/2 * 57 * 57 * sin(117°)

Area ≈ 1/2 * 57 * 57 * 0.927

Area ≈ 1482.135 cm²

c) Angle ABC using Law of Sines:

The Law of Sines states that for any triangle ABC with sides a, b, and c, and opposite angles A, B, and C:

sin(A)/a = sin(B)/b = sin(C)/c

We already know angle BAC (117 degrees) and sides AB and BC. We can use this information to find angle ABC:

sin(ABC) =
arcsin(( BC * sin(BAC))/(AB))

Now, find the angle ABC:

ABC =
arcsin((99.72*sin117)/(57) )

ABC ≈ arcsin(1.221)

ABC ≈ 1.27 radians

ABC ≈ 72.86°

User Martin Le
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