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Zorluk: ZorTriangles: 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}Cevap
  2. B
    1616
  3. 2020Cevap
  4. D
    2222
  5. E
    274\sqrt{274}

Cevap

The possible lengths of side ACAC are 2372\sqrt{37} and 2020.
Using the triangle area formula Area=12absinB\text{Area} = \frac{1}{2} \cdot a \cdot b \cdot \sin B, we find sinB=45\sin B = \frac{4}{5}. Because sinB\sin B is positive in both Quadrant I and Quadrant II, angle BB can be either acute or obtuse. If angle BB is acute, cosB=35\cos B = \frac{3}{5}, giving AC=72+1522(7)(15)(0.6)=237AC = \sqrt{7^2 + 15^2 - 2(7)(15)(0.6)} = 2\sqrt{37}. If angle BB is obtuse, cosB=35\cos B = -\frac{3}{5}, giving AC=72+1522(7)(15)(0.6)=20AC = \sqrt{7^2 + 15^2 - 2(7)(15)(-0.6)} = 20. Both values represent valid triangle configurations.

Adım Adım Çözüm

1
Determine the sine of angle BB using the area formula.
Area=12ABBCsinB    42=12715sinB    sinB=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 12absinθ\frac{1}{2} a b \sin \theta.
2
Find the possible values for cosB\cos B.
Since sinB=45\sin B = \frac{4}{5}, cosB\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+1522(7)(15)(35)=49+225126=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+BC22(AB)(BC)cosBAC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos B with cosB=0.6\cos B = 0.6.
4
Calculate the length of side ACAC when angle BB is obtuse.
AC2=72+1522(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 cosB=0.6\cos B = -0.6.

Anahtar Kavram

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