All Circles Are Similar - Precalculus

  1. Use the Law of Cosines: find cc if a=5a=5, b=5b=5, and included angle C=60∘C=60^\circ.
    c=52+52−2⋅5⋅5cos⁡60∘=5c=\sqrt{5^2+5^2-2\cdot^5\cdot^5\cos 60^\circ}=5. Isosceles with 60°60° angle forms equilateral triangle.
  2. Find AA if sin⁡A=12\sin A=\frac{1}{2} and AA is acute.
    30∘30^\circ. sin⁡30°=12\sin 30°=\frac{1}{2} from special right triangle.
  3. State the Law of Sines for triangle ABCABC using sides a,b,ca,b,c opposite angles A,B,CA,B,C.
    asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}. Relates each side to the sine of its opposite angle.
  4. In any triangle, which side is opposite angle AA when using standard notation?
    aa. Standard notation pairs lowercase sides with uppercase angles.
  5. Use the Law of Sines: if a=10a=10, A=30∘A=30^\circ, and B=60∘B=60^\circ, what is bb?
    b=10sin⁡60∘sin⁡30∘=103b=10\frac{\sin 60^\circ}{\sin 30^\circ}=10\sqrt{3}. Apply asin⁡A=bsin⁡B\frac{a}{\sin A}=\frac{b}{\sin B} and simplify.
  6. Identify the included angle for sides bb and cc in triangle ABCABC (standard opposite-side notation).
    AA. Angle AA is between sides bb and cc in standard notation.
  7. What is the triangle angle sum rule used after finding two angles?
    A+B+C=180∘A+B+C=180^\circ. Sum of interior angles in any triangle equals 180°180°.
  8. Use the Law of Sines: if a=12a=12, A=90∘A=90^\circ, and B=30∘B=30^\circ, what is bb?
    b=12sin⁡30∘sin⁡90∘=6b=12\frac{\sin 30^\circ}{\sin 90^\circ}=6. Apply sine ratio with sin⁡90∘=1\sin 90^\circ=1 and sin⁡30∘=12\sin 30^\circ=\frac{1}{2}.
  9. What is the area formula using two sides and the included angle, for sides b,cb,c and angle AA?
    K=12bcsin⁡AK=\frac{1}{2}bc\sin A. Uses half the product of two sides times sine of included angle.
  10. Find the magnitude of the resultant of two forces 1010 and 1010 with included angle 60∘60^\circ.
    R=102+102+2⋅10⋅10cos⁡60∘=103R=\sqrt{10^2+10^2+2\cdot10\cdot10\cos60^\circ}=10\sqrt{3}. Use cosine formula with angle between forces.
  11. Which triangle data type is directly solvable using the Law of Sines without extra steps: AASAAS, ASAASA, SASSAS, or SSSSSS?
    AASAAS or ASAASA. Both have two angles and one side, perfect for sine ratio.
  12. Use the Law of Sines: if a=8a=8, A=45∘A=45^\circ, and B=30∘B=30^\circ, what is bb (exact form)?
    b=8sin⁡30∘sin⁡45∘=42b=8\frac{\sin 30^\circ}{\sin 45^\circ}=4\sqrt{2}. Apply sine ratio and simplify using sin⁡30∘=12\sin 30^\circ=\frac{1}{2}, sin⁡45∘=22\sin 45^\circ=\frac{\sqrt{2}}{2}.
  13. State the Law of Cosines formula for angle AA using sides a,b,ca,b,c.
    cos⁡A=b2+c2−a22bc\cos A=\frac{b^2+c^2-a^2}{2bc}. Rearranges Law of Cosines to solve for angle's cosine.
  14. What triangle information type is the standard use case for the Law of Sines: AAS,ASA,SAA,SSA,AAS, ASA, SAA, SSA, or SSSSSS?
    AAS,ASA,or SAAAAS, ASA, \text{or } SAA. Two angles and a side allow unique triangle solution.
  15. State the Law of Sines for a triangle with sides a,b,ca,b,c opposite angles A,B,CA,B,C.
    asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}. Relates ratios of sides to sines of opposite angles.
  16. State the Law of Cosines formula that solves for side aa in triangle ABCABC.
    a2=b2+c2−2bccos⁡Aa^2=b^2+c^2-2bc\cos A. Relates side aa to the other sides and angle AA.
  17. Which triangle data type is the classic ambiguous case for the Law of Sines: SSASSA, SASSAS, or SSSSSS?
    SSASSA. Two sides and non-included angle can yield 0, 1, or 2 triangles.
  18. Classify the triangle by angles if a=3a=3, b=4b=4, and c=5c=5 (largest side is 55).
    Right, since 52=32+42\text{Right, since }5^2=3^2+4^2. Pythagorean theorem confirms 90°90° angle.
  19. Find the missing angle if A=35∘A=35^\circ and B=75∘B=75^\circ in a triangle.
    C=70∘C=70^\circ. Triangle angles sum to 180∘180^\circ.
  20. State the Law of Cosines formula for side aa in terms of b,c,b,c, and included angle AA.
    a2=b2+c2−2bccos⁡Aa^2=b^2+c^2-2bc\cos A. Generalizes Pythagorean theorem with cosine correction term.
  21. In the SSASSA case, what is the first step to test for 00, 11, or 22 triangles when angle AA is known?
    Compute h=bsin⁡Ah=b\sin A. Height hh determines if side aa can reach the opposite side.
  22. In the SSASSA case with acute AA, what condition gives two triangles?
    h<bh<b and h<a<bh<a<b where h=bsin⁡Ah=b\sin A. Side aa can swing to two positions when AA is acute.
  23. Use the Law of Sines: find bb if A=45∘A=45^\circ, B=30∘B=30^\circ, and a=8a=8.
    b=8sin⁡30∘sin⁡45∘=42b=\frac{8\sin 30^\circ}{\sin 45^\circ}=4\sqrt{2}. Calculate: b=8(0.5)22=42b=\frac{8(0.5)}{\frac{\sqrt{2}}{2}}=4\sqrt{2}.
  24. Find CC if A=35∘A=35^\circ and B=65∘B=65^\circ in a triangle.
    80∘80^\circ. Apply angle sum: C=180°−35°−65°=80°C=180°-35°-65°=80°.
  25. In the SSASSA case with given AA, aa, and bb, what condition gives no triangle?
    a<h=bsin⁡Aa<h=b\sin A. Side aa is too short to reach the opposite side.
  26. What triangle information type is the standard use case for the Law of Cosines: SAS,SSS,SAS, SSS, or ASAASA?
    SAS or SSSSAS\text{ or }SSS. Two sides with included angle or three sides determine triangle.
  27. State the altitude relation for the SSASSA case with known A,a,A,a, and adjacent side bb.
    h=bsin⁡Ah=b\sin A. Height from vertex to opposite side uses sine of angle.
  28. For SSASSA with A=30∘A=30^\circ, a=4a=4, b=10b=10, how many triangles are possible?
    0 triangles0\text{ triangles}. Since a<bsin⁡A=5a<b\sin A=5, no triangle exists.
  29. State the Law of Sines form that directly solves for aa given A,b,A,b, and BB.
    a=bsin⁡Asin⁡Ba=\frac{b\sin A}{\sin B}. Cross-multiply Law of Sines to isolate desired side.
  30. State the Law of Cosines formula that solves for angle AA given sides a,b,ca,b,c.
    cos⁡A=b2+c2−a22bc\cos A=\frac{b^2+c^2-a^2}{2bc}. Rearranges to solve for the cosine of angle AA.
  31. Identify the ambiguous case in triangle solving that can yield 0,1,0,1, or 22 solutions.
    SSASSA. Two sides and non-included angle may have multiple solutions.
  32. For SSASSA with A=30∘A=30^\circ, a=6a=6, b=10b=10, how many triangles are possible?
    2 triangles2\text{ triangles}. Since bsin⁡A<a<bb\sin A<a<b, two triangles possible.

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