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Solve the following triangles using the Sine Rule and angle sum property of a triangle. Round answers to the nearest whole number.

Mathematics

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Here are the solutions for the given triangles using the Sine Rule and the angle sum property of a triangle, with answers rounded to the nearest whole number.

The Sine Rule states:

asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

The sum of angles in a triangle is 180∘180^\circ:

A+B+C=180∘A + B + C = 180^\circ

1) Given: a=24.5 cma = 24.5 \text{ cm}, B=44∘B = 44^\circ, C=76∘C = 76^\circ

  1. Find angle AA.

    Using the angle sum property:

    A=180∘−B−CA = 180^\circ - B - C A=180∘−44∘−76∘A = 180^\circ - 44^\circ - 76^\circ A=180∘−120∘A = 180^\circ - 120^\circ A=60∘A = 60^\circ
  2. Find side bb.

    Using the Sine Rule:

    bsin⁡B=asin⁡A\frac{b}{\sin B} = \frac{a}{\sin A} b=asin⁡Bsin⁡Ab = \frac{a \sin B}{\sin A} b=24.5sin⁡44∘sin⁡60∘b = \frac{24.5 \sin 44^\circ}{\sin 60^\circ} b≈24.5×0.69470.8660b \approx \frac{24.5 \times 0.6947}{0.8660} b≈19.65 cmb \approx 19.65 \text{ cm}

    Rounding to the nearest whole number:

    b≈20 cmb \approx \boxed{20 \text{ cm}}
  3. Find side cc.

    Using the Sine Rule:

    csin⁡C=asin⁡A\frac{c}{\sin C} = \frac{a}{\sin A} c=asin⁡Csin⁡Ac = \frac{a \sin C}{\sin A} c=24.5sin⁡76∘sin⁡60∘c = \frac{24.5 \sin 76^\circ}{\sin 60^\circ} c≈24.5×0.97030.8660c \approx \frac{24.5 \times 0.9703}{0.8660} c≈27.45 cmc \approx 27.45 \text{ cm}

    Rounding to the nearest whole number:

    c≈27 cmc \approx \boxed{27 \text{ cm}}

    2) Given: A=62∘A = 62^\circ, b=14 mb = 14 \text{ m}, C=70.75∘C = 70.75^\circ

  4. Find angle BB.

    Using the angle sum property:

    B=180∘−A−CB = 180^\circ - A - C B=180∘−62∘−70.75∘B = 180^\circ - 62^\circ - 70.75^\circ B=180∘−132.75∘B = 180^\circ - 132.75^\circ B=47.25∘B = 47.25^\circ

    Rounding to the nearest whole number:

    B≈47∘B \approx \boxed{47^\circ}
  5. Find side aa.

    Using the Sine Rule:

    asin⁡A=bsin⁡B\frac{a}{\sin A} = \frac{b}{\sin B} a=bsin⁡Asin⁡Ba = \frac{b \sin A}{\sin B} a=14sin⁡62∘sin⁡47.25∘a = \frac{14 \sin 62^\circ}{\sin 47.25^\circ} a≈14×0.88290.7342a \approx \frac{14 \times 0.8829}{0.7342} a≈16.84 ma \approx 16.84 \text{ m}

    Rounding to the nearest whole number:

    a≈17 ma \approx \boxed{17 \text{ m}}
  6. Find side cc.

    Using the Sine Rule:

    csin⁡C=bsin⁡B\frac{c}{\sin C} = \frac{b}{\sin B} c=bsin⁡Csin⁡Bc = \frac{b \sin C}{\sin B} c=14sin⁡70.75∘sin⁡47.25∘c = \frac{14 \sin 70.75^\circ}{\sin 47.25^\circ} c≈14×0.94420.7342c \approx \frac{14 \times 0.9442}{0.7342} c≈18.00 mc \approx 18.00 \text{ m}

    Rounding to the nearest whole number:

    c≈18 mc \approx \boxed{18 \text{ m}}

    3) Given: A=31∘A = 31^\circ, B=65∘B = 65^\circ, c=16 mc = 16 \text{ m}

  7. Find angle CC.

    Using the angle sum property:

    C=180∘−A−BC = 180^\circ - A - B C=180∘−31∘−65∘C = 180^\circ - 31^\circ - 65^\circ C=180∘−96∘C = 180^\circ - 96^\circ C=84∘C = 84^\circ
  8. Find side aa.

    Using the Sine Rule:

    asin⁡A=csin⁡C\frac{a}{\sin A} = \frac{c}{\sin C} a=csin⁡Asin⁡Ca = \frac{c \sin A}{\sin C} a=16sin⁡31∘sin⁡84∘a = \frac{16 \sin 31^\circ}{\sin 84^\circ} a≈16×0.51500.9945a \approx \frac{16 \times 0.5150}{0.9945} a≈8.29 ma \approx 8.29 \text{ m}

    Rounding to the nearest whole number

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