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Difficulty: HardTriangles: Properties, Perimeter, and Area

In triangle ABCABC, the length of side ABAB is 77 and the length of side BCBC is 1515. If the area of triangle ABCABC is 4242, which of the following could be the length of side ACAC? Select all such lengths.

  1. 2372\sqrt{37}Answer
  2. B
    1616
  3. 2020Answer
  4. D
    2222
  5. E
    274\sqrt{274}

Answer

The possible lengths of side ACAC are 2372\sqrt{37} and 2020.
Using the triangle area formula Area=12⋅a⋅b⋅sin⁡B\text{Area} = \frac{1}{2} \cdot a \cdot b \cdot \sin B, we find sin⁡B=45\sin B = \frac{4}{5}. Because sin⁡B\sin B is positive in both Quadrant I and Quadrant II, angle BB can be either acute or obtuse. If angle BB is acute, cos⁡B=35\cos B = \frac{3}{5}, giving AC=72+152−2(7)(15)(0.6)=237AC = \sqrt{7^2 + 15^2 - 2(7)(15)(0.6)} = 2\sqrt{37}. If angle BB is obtuse, cos⁡B=−35\cos B = -\frac{3}{5}, giving AC=72+152−2(7)(15)(−0.6)=20AC = \sqrt{7^2 + 15^2 - 2(7)(15)(-0.6)} = 20. Both values represent valid triangle configurations.

Step-by-Step Solution

1
Determine the sine of angle BB using the area formula.
Area=12⋅AB⋅BC⋅sin⁡B  ⟹  42=12⋅7⋅15⋅sin⁡B  ⟹  sin⁡B=84105=45\text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin B \implies 42 = \frac{1}{2} \cdot 7 \cdot 15 \cdot \sin B \implies \sin B = \frac{84}{105} = \frac{4}{5}.
The area of a triangle with two given sides and an included angle is 12absin⁡θ\frac{1}{2} a b \sin \theta.
2
Find the possible values for cos⁡B\cos B.
Since sin⁡B=45\sin B = \frac{4}{5}, cos⁡B\cos B can be either 35\frac{3}{5} (if angle BB is acute) or −35-\frac{3}{5} (if angle BB is obtuse).
Sine is positive in both the first and second quadrants, permitting both acute and obtuse angles for triangle ABCABC.
3
Calculate the length of side ACAC when angle BB is acute.
AC2=72+152−2(7)(15)(35)=49+225−126=148  ⟹  AC=148=237AC^2 = 7^2 + 15^2 - 2(7)(15)\left(\frac{3}{5}\right) = 49 + 225 - 126 = 148 \implies AC = \sqrt{148} = 2\sqrt{37}.
Apply the Law of Cosines: AC2=AB2+BC2−2(AB)(BC)cos⁡BAC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos B with cos⁡B=0.6\cos B = 0.6.
4
Calculate the length of side ACAC when angle BB is obtuse.
AC2=72+152−2(7)(15)(−35)=49+225+126=400  ⟹  AC=400=20AC^2 = 7^2 + 15^2 - 2(7)(15)\left(-\frac{3}{5}\right) = 49 + 225 + 126 = 400 \implies AC = \sqrt{400} = 20.
Apply the Law of Cosines with cos⁡B=−0.6\cos B = -0.6.

Key Concept

Triangle Area via Included Angle and Dual Solutions in Non-Right Triangles
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Topics

Triangles: Properties, Perimeter, and Area

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