All Circles Are Similar - Geometry

1. State the Law of Cosines formula for finding side $b$ in triangle $ABC$.

   $b^2=a^2+c^2-2ac\cos B$. Relates side $b$ to other sides and included angle $B$.

2. Which method is most appropriate first for an $SAS$ triangle: Law of Sines or Law of Cosines?

   Law of Cosines. Two sides and included angle need the cosine formula.

3. Find $h$ (altitude to side $c$) if $c=20$ and $A=30^\circ$ in the $SSA$ setup.

   $h=c\sin A=20\sin 30^\circ=10$. Height calculation for ambiguous case analysis.

4. What condition yields exactly one triangle in the $SSA$ case when the given angle $A$ is obtuse?

   $a>c$. Obtuse angle requires opposite side to be longest.

5. Identify whether $a=6$, $c=12$, $A=30^\circ$ gives $0$, $1$, or $2$ triangles using $h=c\sin A$.

   $1$ triangle (right) because $h=6$ and $a=h$. Side equals height, creating perpendicular configuration.

6. What is the common labeling convention relating $a$ and $A$ in triangle $ABC$?

   $a$ is opposite $A$. Standard notation: lowercase sides opposite uppercase angles.

7. What condition guarantees exactly one right triangle in the $SSA$ case when $A$ is acute and $a$ is given?

   $a=h$ where $h=c\sin A$. Side exactly equals height, forming right triangle.

8. Find $b$ using the Law of Cosines if $a=4$, $c=11$, and $B=60^\circ$.

   $b=\sqrt{4^2+11^2-2\cdot^4\cdot11\cos 60^\circ}=\sqrt{93}$. Apply cosine formula with given angle and sides.

9. Identify the second possible angle $B$ in the $SSA$ case if $\sin B=0.5$ and $B$ is not acute.

   $B=180^\circ-30^\circ=150^\circ$. Supplementary angle when sine has two solutions.

10. State the Law of Cosines formula for finding angle $A$ from sides $a$, $b$, and $c$.

    $\cos A=\frac{b^2+c^2-a^2}{2bc}$. Rearranged Law of Cosines to solve for angle $A$.

11. Find $C$ (nearest degree) if $A=40^\circ$, $a=12$, and $c=9$ using Law of Sines.

    $C\approx\sin^{-1}\!\left(\frac{9\sin 40^\circ}{12}\right)\approx 29^\circ$. Use inverse sine with the established proportion.

12. Identify the special simplification of Law of Cosines when the included angle is $90^\circ$.

    $a^2=b^2+c^2$ when $A=90^\circ$. Reduces to Pythagorean theorem when angle is right.

13. State the Law of Cosines formula for finding angle $B$ from sides $a$, $b$, and $c$.

    $\cos B=\frac{a^2+c^2-b^2}{2ac}$. Rearranged Law of Cosines to solve for angle $B$.

14. State the Law of Sines for triangle $ABC$ with sides $a$, $b$, $c$ opposite angles $A$, $B$, $C$.

    $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$. Relates ratios of sides to sines of opposite angles.

15. Find the third side $c$ of a triangle if $a=9$, $b=9$, and $C=60^\circ$.

    $c=\sqrt{9^2+9^2-2\cdot^9\cdot^9\cos 60^\circ}=9$. Isosceles triangle with 60° creates equilateral result.

16. Find the resultant magnitude $R$ of two forces $10$ and $6$ with included angle $60^\circ$.

    $R=\sqrt{10^2+6^2+2\cdot10\cdot^6\cos 60^\circ}=\sqrt{196}=14$. Vector addition using cosine formula for resultant.

17. What restriction must be checked after computing $B_2=180^\circ-B_1$ in the $SSA$ case?

    $A+B_2<180^\circ$. Ensures the triangle angle sum remains valid.

18. What is the ambiguous case configuration for the Law of Sines that can yield $0$, $1$, or $2$ triangles?

    $SSA$. Two sides and non-included angle can create ambiguity.

19. Find $B$ using the Law of Cosines if $a=6$, $b=10$, and $c=8$ (nearest degree).

    $B\approx\cos^{-1}\!\left(\frac{6^2+8^2-10^2}{2\cdot^6\cdot^8}\right)\approx 90^\circ$. Substitute into angle formula with sides 6, 10, 8.

20. Identify the missing step: after using Law of Sines to find $B$, how do you get the other possible $B$?

    $B_2=180^\circ-B_1$. Supplementary angles share the same sine value.

21. Which method is most appropriate first for an $ASA$ or $AAS$ triangle: Law of Sines or Law of Cosines?

    Law of Sines. Two angles and any side allow direct use of sine ratios.

22. Find $C$ if $A=35^\circ$ and $B=80^\circ$.

    $C=180^\circ-35^\circ-80^\circ=65^\circ$. Apply angle sum property of triangles.

23. What is the triangle angle-sum equation you must use after finding two angles with trig laws?

    $A+B+C=180^\circ$. All triangle angles must sum to 180 degrees.

24. Find $c$ using the Law of Cosines if $a=5$, $b=8$, and $C=90^\circ$.

    $c=\sqrt{5^2+8^2}=\sqrt{89}$. Right angle makes $\cos 90° = 0$ in the formula.

25. Find the resultant magnitude $R$ of two forces $8$ and $5$ with included angle $120^\circ$.

    $R=\sqrt{8^2+5^2+2\cdot^8\cdot^5\cos 120^\circ}=7$. Apply resultant formula with obtuse included angle.

26. What condition yields no triangle in the $SSA$ case when the given angle $A$ is obtuse?

    $a\le c$. Obtuse angle cannot have shorter opposite side.

27. Find $a$ using the Law of Sines if $b=10$, $B=30^\circ$, and $A=45^\circ$.

    $a=10\frac{\sin 45^\circ}{\sin 30^\circ}=10\sqrt{2}$. Cross-multiply and substitute known angle values.

28. Find $b$ if $a=18$, $A=50^\circ$, and $B=80^\circ$ (nearest tenth).

    $b\approx 18\frac{\sin 80^\circ}{\sin 50^\circ}\approx 23.1$. Cross-multiply the sine proportion to solve.

29. Identify whether Law of Sines or Law of Cosines is best first for given $a$, $b$, and included angle $C$.

    Law of Cosines. Two sides with included angle need cosine approach.

30. What condition yields exactly one triangle in the $SSA$ case when $A$ is acute and the opposite side is long enough?

    $a\ge c$. Opposite side is longer than adjacent, ensuring uniqueness.

31. Identify whether $a=14$, $c=12$, $A=30^\circ$ gives $0$, $1$, or $2$ triangles.

    $1$ triangle because $a\ge c$. Opposite side is longer than adjacent side.

32. State the Law of Cosines form typically used for resultant vectors with included angle $\theta$.

    $R^2=p^2+q^2+2pq\cos\theta$. General form for vector resultant calculations.

33. Find the area if $b=10$, $c=12$, and $A=30^\circ$.

    $\text{Area}=\frac{1}{2}(10)(12)\sin 30^\circ=30$. Apply area formula with sine of included angle.

34. Identify the correct proportion to solve for $b$ if $a$, $A$, $B$ are known using Law of Sines.

    $\frac{b}{\sin B}=\frac{a}{\sin A}$. Standard proportion setup for unknown side calculation.

35. Identify whether $a=5$, $c=12$, $A=30^\circ$ gives $0$, $1$, or $2$ triangles using $h=c\sin A$.

    $0$ triangles because $h=6$ and $5<6$. Side 5 is shorter than required height 6.

36. Find $\sin B$ if $a=9$, $A=30^\circ$, and $b=12$ using Law of Sines.

    $\sin B=\frac{b\sin A}{a}=\frac{12\sin 30^\circ}{9}=\frac{2}{3}$. Set up proportion to find sine of unknown angle.

37. Identify whether $a=9$, $c=12$, $A=30^\circ$ gives $0$, $1$, or $2$ triangles using $h=c\sin A$.

    $2$ triangles because $h=6$ and $6<9<12$. Height is 6; side 9 is between height and adjacent side.

38. Find $B$ if $b=15$, $a=10$, and $c=12$ using $\cos B$ (nearest degree).

    $B\approx\cos^{-1}\!\left(\frac{10^2+12^2-15^2}{2\cdot10\cdot12}\right)\approx 104^\circ$. Apply cosine angle formula with three known sides.

39. State the Law of Cosines formula for finding side $a$ in triangle $ABC$.

    $a^2=b^2+c^2-2bc\cos A$. Relates side $a$ to other sides and included angle $A$.

40. Find $c$ if $a=13$, $b=14$, and $C=60^\circ$ using the Law of Cosines (nearest tenth).

    $c\approx\sqrt{13^2+14^2-2\cdot13\cdot14\cos 60^\circ}\approx 13.5$. Apply cosine formula with two sides and included angle.

41. State the Law of Cosines formula for finding side $c$ in triangle $ABC$.

    $c^2=a^2+b^2-2ab\cos C$. Relates side $c$ to other sides and included angle $C$.

42. Find $B$ if $a=12$, $A=40^\circ$, and $b=9$ using the Law of Sines (nearest degree).

    $B\approx\sin^{-1}\!\left(\frac{9\sin 40^\circ}{12}\right)\approx 29^\circ$. Use inverse sine after setting up the proportion.

43. Find $A$ using the Law of Cosines if $a=10$, $b=7$, and $c=9$ (nearest degree).

    $A\approx\cos^{-1}\!\left(\frac{7^2+9^2-10^2}{2\cdot7\cdot9}\right)\approx 74^\circ$. Use inverse cosine after substituting side values.

44. Identify whether Law of Sines or Law of Cosines is best first for given $A$, $B$, and side $a$.

    Law of Sines. Two angles and one side use sine relationships.

45. Find the included angle $C$ if $a=7$, $b=10$, and $c=13$ (nearest degree).

    $C\approx\cos^{-1}\!\left(\frac{7^2+10^2-13^2}{2\cdot7\cdot10}\right)\approx 105^\circ$. Use inverse cosine with three known sides.

46. Find $a$ if $\frac{a}{\sin 50^\circ}=\frac{14}{\sin 70^\circ}$ (nearest tenth).

    $a\approx 14\frac{\sin 50^\circ}{\sin 70^\circ}\approx 11.4$. Cross-multiply and divide to isolate unknown side.

47. What condition yields two possible triangles in the $SSA$ case when $A$ is acute and adjacent side is $c$?

    $h<a<c$ where $h=c\sin A$. Side length allows two possible triangle configurations.

48. Find $a$ using the Law of Cosines if $b=7$, $c=9$, and $A=60^\circ$.

    $a=\sqrt{7^2+9^2-2\cdot^7\cdot^9\cos 60^\circ}=\sqrt{67}$. Substitute values into cosine formula and simplify.

49. What condition guarantees no triangle in the $SSA$ case when given $A$ (acute), $a$, and adjacent side $c$?

    $a<h$ where $h=c\sin A$. Side opposite acute angle is too short to reach.

50. Identify the key calculator mode needed when using $\sin^{-1}$ or $\cos^{-1}$ with degrees.

    Degree mode. Calculator must be set for degree calculations.

51. What is the area formula using two sides and the included angle, useful with trig-law problems?

    $\text{Area}=\frac{1}{2}bc\sin A$. Uses two sides and included angle for area.

52. Which method is most appropriate first for an $SSS$ triangle: Law of Sines or Law of Cosines?

    Law of Cosines. Three sides require the cosine formula approach.

53. Find $A$ if $a=8$, $b=5$, and $c=7$ using $\cos A$ (nearest degree).

    $A\approx\cos^{-1}\!\left(\frac{5^2+7^2-8^2}{2\cdot^5\cdot^7}\right)\approx 79^\circ$. Use inverse cosine formula for finding angles.

54. State the Law of Cosines formula for finding angle $C$ from sides $a$, $b$, and $c$.

    $\cos C=\frac{a^2+b^2-c^2}{2ab}$. Rearranged Law of Cosines to solve for angle $C$.

55. Find $c$ using the Law of Sines if $a=8$, $A=60^\circ$, and $C=45^\circ$.

    $c=8\frac{\sin 45^\circ}{\sin 60^\circ}=\frac{8\sqrt{6}}{3}$. Set up proportion and solve for unknown side.

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