Answer :

To solve triangle ABC with [tex]B = 35^\circ 00'[/tex], [tex]a = 38.5[/tex], and [tex]b = 30.9[/tex], we will use the Law of Sines, which states that for a triangle with sides [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex], opposite their respective angles [tex]A[/tex], [tex]B[/tex], and [tex]C[/tex]:

[tex]\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}[/tex]

  1. Calculate Angle A:

    We have [tex]\sin B = \sin 35^\circ[/tex], and from the Law of Sines:

    [tex]\frac{38.5}{\sin A} = \frac{30.9}{\sin 35^\circ}[/tex]

    Solving for [tex]\sin A[/tex]:

    [tex]\sin A = \frac{38.5 \times \sin 35^\circ}{30.9}[/tex]

    Calculating [tex]\sin 35^\circ[/tex] first, and then substitution gives:

    [tex]\sin A \approx \frac{38.5 \times 0.5736}{30.9} \approx 0.7147[/tex]

    Now find [tex]A[/tex] using [tex]\arcsin(0.7147)[/tex]:

    [tex]A \approx 45.5^\circ[/tex]

  2. Calculate Angle C:

    Since the sum of angles in a triangle is [tex]180^\circ[/tex], we can find [tex]C[/tex] as follows:

    [tex]C = 180^\circ - A - B = 180^\circ - 45.5^\circ - 35^\circ[/tex]

    [tex]C = 99.5^\circ[/tex]

  3. Calculate Side c:

    Using the Law of Sines again:

    [tex]\frac{c}{\sin C} = \frac{a}{\sin A}[/tex]

    Solving for [tex]c[/tex]:

    [tex]c = a \times \frac{\sin C}{\sin A} = 38.5 \times \frac{\sin 99.5^\circ}{\sin 45.5^\circ}[/tex]

    [tex]c \approx 38.5 \times \frac{0.9914}{0.7147} \approx 53.4[/tex]

Therefore, the solution for the triangle is:

  • [tex]A \approx 45.5^\circ[/tex]
  • [tex]C \approx 99.5^\circ[/tex]
  • [tex]c \approx 53.4[/tex]

This completes solving the triangle given the initial data.

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Rewritten by : Barada