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A triangle upper A upper B upper C. The edge upper A upper B is labeled c. The edge upper A upper C is labeled b. The edge upper B upper C is labeled a.

If the area of △ABC=528.3 mm2
, a=21 mm
, and ∠C=57 degrees
, then what is the length of b? Round your answer to the nearest tenth.

(1 point)
b=
mm

1 answer

To find the length of side \( b \) in triangle \( ABC \), we can use the formula for the area of a triangle in terms of two sides and the sine of the included angle:

\[ \text{Area} = \frac{1}{2}ab \sin C \]

Here:

  • The area is given as \( 528.3 , \text{mm}^2 \).
  • Side \( a \) is \( 21 , \text{mm} \).
  • Angle \( C \) is \( 57^\circ \).
  • We want to find side \( b \).

We can rearrange the formula for the area to solve for \( b \):

\[ b = \frac{2 \cdot \text{Area}}{a \cdot \sin C} \]

Now substituting the known values into this formula:

  1. Calculate \( \sin C \):

\[ \sin 57^\circ \approx 0.8387 \]

  1. Plug the values into the equation for \( b \):

\[ b = \frac{2 \cdot 528.3}{21 \cdot 0.8387} \]

  1. Calculate \( 21 \cdot 0.8387 \):

\[ 21 \cdot 0.8387 \approx 17.6137 \]

  1. Now calculate \( b \):

\[ b = \frac{1056.6}{17.6137} \approx 59.97 \]

Finally, rounding \( b \) to the nearest tenth:

\[ b \approx 60.0 , \text{mm} \]

Thus, the length of side \( b \) is \( \boxed{60.0} , \text{mm} \).