- Opposite means across the triangle, not touching the chosen vertex. - A side's opposite angle is at the triangle's remaining vertex.

1. The side across from vertex \(C\) joins the other two vertices, so it is \(AB\). 2. Side \(BC\) lies across from vertex \(A\), so its opposite angle is \(\angle A\).

Side \(AB\); angle \(\angle A\).

- Find the angle across from the known side \(PR\). - Which angle lies across from \(QR\)? - Each side must be paired with its own opposite angle.

1. Side \(PR\) is opposite \(\angle Q\), and side \(QR\) is opposite \(\angle P\). 2. Match each side with its opposite angle: \(\frac{QR}{\sin42^\circ}=\frac{9}{\sin73^\circ}\).

\(\frac{QR}{\sin42^\circ}=\frac{9}{\sin73^\circ}\)

- What operation can you apply to both sides of an equality without changing its truth? - Can you reverse your operation to check that the two forms agree?

1. Both ratios are positive for a nondegenerate triangle, so taking reciprocals preserves equality. 2. The equivalent form is \(\frac{\sin A}{a}=\frac{\sin B}{b}\).

\(\frac{\sin A}{a}=\frac{\sin B}{b}\). It is equivalent because both sides of the original equality were inverted.

- What are the base angles in an isosceles triangle with the given vertex angle? - Match a known side to its opposite angle. - Use that pair to set up the Law of Sines for a congruent side.

1. The base angles are equal: \(\alpha=\beta=\frac{180^\circ-36^\circ}{2}=72^\circ\). 2. By the Law of Sines, \(\frac{a}{\sin(72^\circ)}=\frac{8.4}{\sin(36^\circ)}\). 3. Thus, \(a=\frac{8.4\sin(72^\circ)}{\sin(36^\circ)}\approx13.59\,\text{cm}\). Since the triangle is isosceles, \(b=a\).

\(a\approx13.59\,\text{cm}\), \(b\approx13.59\,\text{cm}\)

- First find the third interior angle. - Which law links a side to the sine of its opposite angle? - Make sure each unknown side is paired with its own opposite angle.

1. The angle sum gives \(\gamma=180^\circ-42^\circ-75^\circ=63^\circ\). 2. By the Law of Sines, \(b=\frac{8.5\sin(75^\circ)}{\sin(42^\circ)}\approx12.27\,\text{cm}\). 3. Similarly, \(c=\frac{8.5\sin(63^\circ)}{\sin(42^\circ)}\approx11.32\,\text{cm}\).

\(\gamma=63^\circ\), \(b\approx12.27\,\text{cm}\), \(c\approx11.32\,\text{cm}\)

- Find the third angle using the triangle angle sum. - Use the known side-angle pair in the Law of Sines. - Keep each side matched with its opposite angle.

1. Find the third angle: \(\gamma=180^\circ-40^\circ-60^\circ=80^\circ\). 2. By the Law of Sines, \(a=\frac{9\sin(40^\circ)}{\sin(80^\circ)}\approx 5.9\,\text{cm}\). 3. Similarly, \(b=\frac{9\sin(60^\circ)}{\sin(80^\circ)}\approx 7.9\,\text{cm}\).

\(\gamma=80^\circ\), \(a\approx 5.9\,\text{cm}\), and \(b\approx 7.9\,\text{cm}\).

- Find the third angle before choosing a side-angle pair. - Match each side with its opposite angle. - Use the known baseline and its opposite angle in the Law of Sines.

1. The third angle is \(\angle C=180^\circ-48^\circ-67^\circ=65^\circ\). 2. Side \(AB=220\,\text{m}\) is opposite \(65^\circ\), and \(AC\) is opposite \(67^\circ\). 3. By the Law of Sines, \(\frac{AC}{\sin(67^\circ)}=\frac{220}{\sin(65^\circ)}\). 4. Therefore, \(AC=\frac{220\sin(67^\circ)}{\sin(65^\circ)}\approx223.446\,\text{m}\), which rounds to \(223\,\text{m}\).

\(AC\approx223\,\text{m}\)

- Identify the angle across from the known side and across from the unknown side. - Keep each side paired with its opposite angle before substituting values.

1. Match each side with its opposite angle: \(\frac{b}{\sin45^\circ}=\frac{6}{\sin30^\circ}\). 2. Therefore \(b=\frac{6\sin45^\circ}{\sin30^\circ}=6\sqrt{2}\,\text{cm}\).

\(\frac{b}{\sin45^\circ}=\frac{6}{\sin30^\circ}\); \(b=6\sqrt{2}\,\text{cm}\).

- Which given side has a known opposite angle? - After making the proportion, check whether the resulting sine could describe a triangle angle.

1. The given opposite pair is \(a,A\). The Law of Sines yields \(\frac{\sin B}{9}=\frac{\sin40^\circ}{5}\). 2. Thus \(\sin B=\frac{9\sin40^\circ}{5}\approx1.157>1\). 3. No angle has sine greater than \(1\), so no such triangle exists.

No. \(\sin B=\frac{9\sin40^\circ}{5}\approx1.157>1\), so the data cannot form a triangle.

- Find the third angle using the triangle angle sum. - Which angle is opposite the known side, and which is opposite \(BC\)? - Keep each side together with its opposite angle in the proportion.

1. The third angle is \(C=180^\circ-50^\circ-60^\circ=70^\circ\). 2. \(BC\) is opposite \(50^\circ\), and \(AB=10\,\text{cm}\) is opposite \(70^\circ\). 3. \(\frac{BC}{\sin50^\circ}=\frac{10}{\sin70^\circ}\), so \(BC=\frac{10\sin50^\circ}{\sin70^\circ}\approx8.2\,\text{cm}\).

\(BC\approx8.2\,\text{cm}\)

- Match side \(a\) with angle \(A\) and side \(b\) with angle \(B\). - What second angle can have the same sine? - Check whether either choice leaves room for a positive third angle.

1. \(\frac{\sin A}{5}=\frac{\sin40^\circ}{7}\), so \(\sin A=\frac{5\sin40^\circ}{7}\approx0.4591\). 2. The acute solution is \(A\approx27.3^\circ\), or \(27^\circ\) to the nearest degree. 3. The supplementary candidate \(152.7^\circ\) together with \(B=40^\circ\) exceeds \(180^\circ\), so it is impossible.

\(A\approx27^\circ\). The supplementary angle would make the angle sum exceed \(180^\circ\).

- Which angle is opposite \(AC\), and which is opposite \(BC\)? - Write the two Law-of-Sines ratios and divide them. - Which of the two known angles is larger?

1. Side \(AC\) is opposite \(B=67^\circ\) and side \(BC\) is opposite \(A=42^\circ\). 2. The Law of Sines gives \(\frac{AC}{BC}=\frac{\sin67^\circ}{\sin42^\circ}\). 3. In the acute-angle range, \(\sin67^\circ>\sin42^\circ\), so \(AC>BC\). Station \(A\) is farther from the buoy.

\(\frac{AC}{BC}=\frac{\sin67^\circ}{\sin42^\circ}\). Station \(A\) is farther from \(C\).

- Sketch each triangle and place the known values. - Which set of information lets you form an equation with just the required unknown?

1. For a), two angles and one side let us find an opposite angle-side pair, so the Law of Sines can find another side. 2. For b), the two sides and their included angle form an SAS situation, so the Law of Cosines finds the third side directly.

a) Law of Sines, because a known side can be matched to its opposite angle. b) Law of Cosines, because two sides and their included angle are known.

- Use the given opposite pair \(a,A\) to calculate \(\sin B\). - After finding one possible angle, is it the only angle with that sine value? - Does each candidate \(B\) fit together with \(A\) in one triangle?

1. \(\frac{\sin B}{8}=\frac{\sin30^\circ}{6}\), so \(\sin B=\frac{8\cdot1/2}{6}=\frac{2}{3}\). 2. One possible angle is \(B\approx41.8^\circ\). The other is \(180^\circ-41.8^\circ\approx138.2^\circ\). 3. Adding \(A=30^\circ\) to either value leaves a positive third angle. Therefore two triangles are possible.

Yes, two triangles. \(B\approx41.8^\circ\) or \(138.2^\circ\), and both leave a positive angle \(C\).

- Do not reject the second angle simply because a calculator shows the first one. - Try each candidate with \(A\) using the triangle angle sum.

1. Both \(42^\circ\) and \(138^\circ\) have approximately the given sine, but \(70^\circ+138^\circ>180^\circ\). 2. Therefore \(B\approx42^\circ\) is the valid choice. 3. \(C\approx180^\circ-70^\circ-42^\circ=68^\circ\).

\(B\approx42^\circ\), \(C\approx68^\circ\). The supplementary value of \(B\) would make the angle sum too large.

- Mark the known measurements on a triangle sketch. - What one equation would connect the known side to the side you need?

1. Side \(a\) is opposite \(A\), and side \(b\) is opposite \(B\). 2. The Law of Sines gives \(\frac{b}{\sin65^\circ}=\frac{7}{\sin40^\circ}\). 3. Thus \(b=\frac{7\sin65^\circ}{\sin40^\circ}\approx9.9\,\text{cm}\).

\(\frac{b}{\sin65^\circ}=\frac{7}{\sin40^\circ}\); \(b\approx9.9\,\text{cm}\).

- SSA is not automatically a congruence criterion. - Use the Law of Sines to find the angle opposite the \(4\,\text{cm}\) side. - Test both the inverse-sine angle and its supplement against the triangle angle sum.

1. For \(\triangle ABC\), apply the Law of Sines: \(\frac{\sin \beta}{4} = \frac{\sin 100^\circ}{7}\). Thus, \(\sin \beta = \frac{4\sin 100^\circ}{7} \approx 0.5627\). 2. One possible value is \(\beta \approx 34.2^\circ\). The supplementary value, approximately \(145.8^\circ\), is impossible because \(100^\circ + 145.8^\circ > 180^\circ\). Therefore, the data determine a unique triangle. 3. The same calculation applies to \(\triangle DEF\), so \(\epsilon \approx 34.2^\circ\), and its supplementary value is also impossible. 4. Both sets determine the same side lengths and angle measures, so the triangles are congruent.

Yes. Each SSA data set produces one unique triangle, so \(\triangle ABC\) and \(\triangle DEF\) are congruent. The second sine solution is impossible because the given \(100^\circ\) angle is already obtuse.

- Identify the angle between sides \(a\) and \(b\). - Use the sum and difference forms of the triangle inequality. - For SSA, solve for \(\sin\beta\) and check an angle and its supplement.

1. The sides \(a\) and \(b\) meet at vertex \(C\), so their included angle is \(\gamma\). Knowing \(\gamma\) would give SAS. 2. The triangle inequality gives \(|6 - 4| < c < 6 + 4\), so \(2\,\text{cm} < c < 10\,\text{cm}\). 3. For c), the Law of Sines gives \(\sin\beta = \frac{6\sin 30^\circ}{4} = 0.75\). 4. Thus, \(\beta \approx 48.6^\circ\) or \(\beta \approx 131.4^\circ\). Both angles can be combined with \(\alpha = 30^\circ\), so two noncongruent triangles are possible.

a) \(\gamma\) b) \(2\,\text{cm} < c < 10\,\text{cm}\) c) The SSA data produce two possible values of \(\beta\), so the triangle is not uniquely determined.

- These are SSA cases, so check for the ambiguous case. - Use the Law of Sines to find the sine of the unknown angle. - A value greater than \(1\) is impossible, and a supplementary angle must still fit the triangle angle sum.

1. Case 1 gives \(b = 8\,\text{cm}\), \(a = 6\,\text{cm}\), and \(\beta = 50^\circ\). By the Law of Sines, \(\sin \alpha = \frac{6\sin 50^\circ}{8} \approx 0.5745\). 2. This gives \(\alpha \approx 35.1^\circ\). The supplementary possibility, \(144.9^\circ\), is impossible because \(144.9^\circ + 50^\circ > 180^\circ\). Therefore, Case 1 produces one triangle. 3. Case 2 gives \(a = 6\,\text{cm}\), \(b = 8\,\text{cm}\), and \(\alpha = 50^\circ\). The Law of Sines would require \(\sin \beta = \frac{8\sin 50^\circ}{6} \approx 1.0214\). 4. Since a sine value cannot exceed \(1\), Case 2 produces no triangle.

Case 1: One triangle Case 2: No triangle

- For Lucas, check whether the given angle is included between the two sides. - For Mia, use the Law of Sines to find a possible value of \(\beta\). - Test whether the supplement of that angle can also fit in the triangle.

1. Lucas has sides \(a\) and \(c\) with their included angle \(\beta\). His triangle is uniquely determined by SAS. 2. Mia has SSA data. By the Law of Sines, \(\sin\beta = \frac{4.5\sin 55^\circ}{7} \approx 0.5266\). 3. One possible angle is \(\beta \approx 31.8^\circ\). Its supplement, \(148.2^\circ\), cannot be used because \(55^\circ + 148.2^\circ > 180^\circ\). 4. Therefore, Mia's measurements also determine exactly one triangle, but this conclusion comes from checking the SSA ambiguous case rather than applying a standard congruence criterion.

Lucas: One unique triangle by SAS. Mia: One unique triangle. The Law of Sines gives only one valid SSA solution.

- Write a Law of Sines equation for \(\sin\beta\). - Remember that an acute angle and its supplement have the same sine. - Check whether each possible angle leaves a positive measure for the third angle.

1. Case 1: By the Law of Sines, \(\sin \beta = \frac{b\sin\alpha}{a} = \frac{5\sin 30^\circ}{8} = 0.3125\). 2. One possible angle is \(\beta \approx 18.2^\circ\). Its supplement, \(161.8^\circ\), cannot be used because \(30^\circ + 161.8^\circ > 180^\circ\). Therefore, exactly one triangle is possible. 3. Case 2: \(\sin \beta = \frac{8\sin 30^\circ}{5} = 0.8\). 4. The two possible angles are \(\beta \approx 53.1^\circ\) and \(\beta \approx 126.9^\circ\). Each can be combined with \(\alpha = 30^\circ\), so two noncongruent triangles are possible.

Case 1: Exactly one triangle is determined. Case 2: Two noncongruent triangles are possible, so the triangle is not uniquely determined.

- Use the Law of Sines to find a missing angle. - Check both an inverse-sine result and its supplement. - Test each candidate with the triangle angle sum.

1. For part a), the Law of Sines gives \(\sin(\gamma)=\frac{5.0\sin(100^\circ)}{8.0}\approx 0.6155\). 2. The possible values are \(\gamma_1\approx 37.99^\circ\) and \(\gamma_2\approx 142.01^\circ\). Since \(100^\circ+142.01^\circ>180^\circ\), the second value cannot form a triangle. Exactly one triangle is possible. 3. For part b), \(\sin(\beta)=\frac{6.0\sin(35^\circ)}{4.5}\approx 0.7648\). 4. The possible values are \(\beta_1\approx 49.89^\circ\) and \(\beta_2\approx 130.11^\circ\). Both satisfy \(35^\circ+\beta<180^\circ\), so two distinct triangles are possible.

a) Exactly one triangle is possible because the supplementary angle would make the angle sum exceed \(180^\circ\). b) Two distinct triangles are possible because both candidate values of \(\beta\) produce a valid third angle.

- Find the third angle before matching opposite sides and angles. - Use the Law of Sines to determine \(c\). - Which given angle lies between the two sides you now know? - Keep the unrounded value of \(c\) when finding the area.

1. The third angle is \(\beta=180^\circ-45^\circ-60^\circ=75^\circ\). 2. By the Law of Sines, \(\frac{c}{\sin(60^\circ)}=\frac{8}{\sin(75^\circ)}\), so \(c=\frac{8\sin(60^\circ)}{\sin(75^\circ)}\approx7.17\,\text{cm}\). 3. The included angle between sides \(b\) and \(c\) is \(\alpha\), so \(A=\frac{1}{2}bc\sin(\alpha)\). 4. Using the unrounded side \(c\), \(A\approx\frac{1}{2}\cdot8\cdot7.1726\cdot\sin(45^\circ)\approx20.29\,\text{cm}^2\).

a) \(c\approx7.17\,\text{cm}\) b) \(A\approx20.29\,\text{cm}^2\)

- Use the Law of Sines to calculate \(\sin(\gamma)\). - A sine value can correspond to two supplementary angles between \(0^\circ\) and \(180^\circ\). - Check the triangle angle sum for both candidates. - Compare \(a\) with the altitude \(c\sin(\alpha)\) and with \(c\).

1. By the Law of Sines, \(\sin(\gamma)=\frac{7\sin(40^\circ)}{5}\approx 0.89990\). 2. The two possible angles are \(\gamma_1=\sin^{-1}(0.89990)\approx 64.15^\circ\) and \(\gamma_2=180^\circ-64.15^\circ\approx 115.85^\circ\). 3. Both are valid because \(40^\circ+64.15^\circ<180^\circ\) and \(40^\circ+115.85^\circ<180^\circ\). 4. Equivalently, the ambiguous-case condition \(c\sin(\alpha)<a<c\) holds: \(7\sin(40^\circ)<5<7\). Therefore, two noncongruent triangles exist.

a) \(\gamma\approx 64.15^\circ\) or \(\gamma\approx 115.85^\circ\) b) Both angles leave a positive third angle. Also, \(c\sin(\alpha)<a<c\), which is the two-triangle condition for this SSA case.

- Use the known opposite side-angle pair in the Law of Sines. - Does the calculator's first angle necessarily describe every possible triangle? - How can the three-angle sum help check a candidate triangle?

1. By the Law of Sines, \(\sin(\beta)=\frac{7.2\sin(38^\circ)}{5.4}\approx0.82088\). 2. The two candidates are \(\beta_1\approx55.17^\circ\) and \(\beta_2\approx124.83^\circ\). 3. Both are valid because each gives \(\beta+38^\circ<180^\circ\). 4. The remaining angles are \(\alpha_1\approx86.83^\circ\) and \(\alpha_2\approx17.17^\circ\).

Two distinct triangles exist. Triangle 1: \(\alpha_1\approx86.83^\circ\), \(\beta_1\approx55.17^\circ\). Triangle 2: \(\alpha_2\approx17.17^\circ\), \(\beta_2\approx124.83^\circ\).

- Find the third angle first. - Use the Law of Sines with the known baseline and its opposite angle. - Match the requested distance with its opposite angle.

1. Find the third angle: \(\gamma=180^\circ-42^\circ-105^\circ=33^\circ\). 2. The distance \(BP\) is opposite \(42^\circ\). By the Law of Sines, \(\frac{BP}{\sin(42^\circ)}=\frac{80}{\sin(33^\circ)}\). 3. Therefore, \(BP=80\frac{\sin(42^\circ)}{\sin(33^\circ)}\approx 98.29\,\text{m}\).

The shelter is approximately \(98.29\,\text{m}\) from point \(B\).

- Find the two congruent base angles first. - Pair each side with its opposite angle in the Law of Sines. - Solve the proportion for the base.

1. The two base angles are congruent, so each is \(\frac{180^\circ-40^\circ}{2}=70^\circ\). 2. By the Law of Sines, \(\frac{c}{\sin(40^\circ)}=\frac{10}{\sin(70^\circ)}\). 3. Therefore, \(c=\frac{10\sin(40^\circ)}{\sin(70^\circ)}\approx 6.84\,\text{cm}\).

\(c\approx 6.84\,\text{cm}\)

- Find the third angle of the triangle first. - Match each side with its opposite angle in the Law of Sines. - Solve the resulting proportion for \(x\).

1. The third angle is \(180^\circ-45^\circ-105^\circ=30^\circ\). 2. By the Law of Sines, \(\frac{x}{\sin(45^\circ)}=\frac{2}{\sin(30^\circ)}\). 3. Thus, \(x=\frac{2\sin(45^\circ)}{\sin(30^\circ)}=2\sqrt{2}\approx 2.83\).

\(x=2\sqrt{2}\approx 2.83\)

- For each situation, list the structural information you know before choosing any formula: right angle, opposite side-angle pair, or two sides with the angle between them. - Prefer a method that can use the displayed givens immediately rather than one that first requires finding extra measurements. - Solve only after you have matched each information pattern to its most direct method.

1. Situation I is a right triangle with an acute angle and adjacent/opposite legs, so right-triangle trigonometry is most direct. 2. Situation II supplies a known side-angle opposite pair and another angle with its opposite side unknown, so the Law of Sines is most direct. 3. Let \(x\) be the side opposite \(71^\circ\). Then \(\frac{x}{\sin(71^\circ)}=\frac{18}{\sin(63^\circ)}\). 4. Thus, \(x=18\frac{\sin(71^\circ)}{\sin(63^\circ)}\approx19.101\), so \(x\approx19.1\). 5. Situation III gives two sides and their included angle, so the Law of Cosines is most direct.

Situation II uses the Law of Sines, and the requested side is \(\approx19.1\). Situation I is most directly solved with right-triangle trigonometry; situation III is most directly solved with the Law of Cosines.

- The perimeter connects all three unknown sides. - What does the Law of Sines tell you about the ratios among those sides? - Is the side opposite the smallest angle your shortest length?

1. The third angle is \(180^\circ-35^\circ-70^\circ=75^\circ\). 2. By the Law of Sines, the three side lengths are \(k\sin35^\circ\), \(k\sin70^\circ\), and \(k\sin75^\circ\) for one positive number \(k\). 3. The perimeter gives \(30=k(\sin35^\circ+\sin70^\circ+\sin75^\circ)\). 4. The requested side is \(\frac{30\sin35^\circ}{\sin35^\circ+\sin70^\circ+\sin75^\circ}\approx6.9\,\text{ft}\).

\(\frac{30\sin35^\circ}{\sin35^\circ+\sin70^\circ+\sin75^\circ}\approx6.9\,\text{ft}\)

- Sketch a compass cross at each survey point to interpret the bearings. - Work out the interior angles the two sight lines make with the east-west road. - Once you know all three angles, which side is opposite the measured baseline?

1. Name the tower \(C\). Since \(AB\) points east, the interior angle at \(A\) is \(90^\circ-40^\circ=50^\circ\). 2. The bearing \(305^\circ\) points \(55^\circ\) west of north. Thus angle \(B=90^\circ-55^\circ=35^\circ\). 3. Angle \(C=180^\circ-50^\circ-35^\circ=95^\circ\). 4. By the Law of Sines, \(\frac{AC}{\sin35^\circ}=\frac{80}{\sin95^\circ}\), giving \(AC\approx46\,\text{m}\).

\(AC\approx46\,\text{m}\)

- Which two given sides are opposite the known and unknown angles? - Use the Law of Sines to find one possible value of \(\alpha\). - Can different angle measures have the same sine? How could that affect the triangle? - Keep calculations unrounded until the end of each possible triangle.

1. This is the ambiguous SSA case. Because \(a\sin(\gamma)<c<a\), two triangles are possible. 2. By the Law of Sines, \(\sin(\alpha)=\frac{6.4\sin(35^\circ)}{4.5}\approx0.81575\). 3. The two angle choices are \(\alpha_1\approx54.66^\circ\) and \(\alpha_2\approx125.34^\circ\). 4. The corresponding remaining angles are \(\beta_1\approx90.34^\circ\) and \(\beta_2\approx19.66^\circ\). 5. Using unrounded angles in the Law of Sines gives \(b_1=\frac{4.5\sin(\beta_1)}{\sin(35^\circ)}\approx7.85\,\text{cm}\) and \(b_2\approx2.64\,\text{cm}\).

Two triangles are possible. Triangle 1: \(\alpha_1\approx54.66^\circ\), \(\beta_1\approx90.34^\circ\), \(b_1\approx7.85\,\text{cm}\). Triangle 2: \(\alpha_2\approx125.34^\circ\), \(\beta_2\approx19.66^\circ\), \(b_2\approx2.64\,\text{cm}\).

- Recognize that the information forms an SSA case. - Use the Law of Sines to find the angle opposite the \(5\,\text{cm}\) side. - Test both inverse-sine possibilities with the \(100^\circ\) angle.

1. The data are \(b = 9\,\text{cm}\), \(c = 5\,\text{cm}\), and \(\beta = 100^\circ\), with the given angle opposite the longer side. 2. Use the Law of Sines: \(\sin \gamma = \frac{5\sin 100^\circ}{9} \approx 0.5471\). 3. One possible angle is \(\gamma \approx 33.2^\circ\). Its supplement, approximately \(146.8^\circ\), is impossible because \(100^\circ + 146.8^\circ > 180^\circ\). 4. Therefore, exactly one triangle satisfies the data. Lara’s conclusion is correct. Tom is incorrect because an obtuse angle does not have to be the included angle for the SSA data to determine a unique triangle.

Lara is correct: the data determine exactly one triangle. Tom’s claim is false; the Law of Sines shows that the only valid additional angle is approximately \(33.2^\circ\).

- Treat \(\triangle ABD\) as an SSA ambiguous case. - Use the Law of Sines and test both inverse-sine angles. - Once the shape of \(\triangle ABD\) is fixed, how many sides of \(BD\) can contain the SSS triangle \(BCD\)?

1. In \(\triangle ABD\), \(BD=5\,\text{cm}\) is opposite \(\angle A=30^\circ\), and \(AB=8\,\text{cm}\). 2. By the Law of Sines, \(\sin\angle D=\frac{8\sin30^\circ}{5}=0.8\). Therefore, \(\angle D\) can be approximately \(53.1^\circ\) or \(126.9^\circ\). Both fit with \(\angle A=30^\circ\), so there are two noncongruent possibilities for \(\triangle ABD\). 3. For either triangle \(ABD\), triangle \(BCD\) has side lengths \(4\,\text{cm}\), \(5\,\text{cm}\), and \(5\,\text{cm}\), so it is determined by SSS up to reflection across \(BD\). 4. For each of the two noncongruent choices for \(\triangle ABD\), the SSS triangle \(BCD\) can be attached on either side of \(BD\). Thus, there are \(2\cdot2=4\) quadrilaterals up to congruence.

1. There are two noncongruent possibilities for \(\triangle ABD\) because the SSA data produce two valid triangles. 2. There are \(4\) quadrilaterals up to congruence: two choices for \(\triangle ABD\), and for each one, two choices for which side of \(BD\) contains \(C\).

- What is \(\sin(90^\circ)\)? - Rearrange the proportion to isolate \(\sin(\alpha)\). - Equal angles have equal sine values.

1. Since \(\gamma=90^\circ\), \(\sin(\gamma)=1\). Therefore, \(\frac{a}{\sin(\alpha)}=\frac{c}{1}=c\), which rearranges to \(\sin(\alpha)=\frac{a}{c}\). 2. The Law of Sines is a general relationship. In a right triangle, it reduces exactly to the right-triangle definition because \(c\) is the hypotenuse and \(a\) is opposite \(\alpha\). 3. If \(\alpha=\beta\), then \(\sin(\alpha)=\sin(\beta)\). From \(\frac{a}{\sin(\alpha)}=\frac{b}{\sin(\beta)}\), it follows that \(a=b\).

1. \(\sin(\alpha)=\frac{a}{c}\) 2. The Law of Sines includes the right-triangle sine ratio as a special case, so there is no contradiction. 3. \(a=b\), which matches the equal-side property of an isosceles triangle.

- Consider the shortest possible segment from the endpoint of side \(c\) to the ray forming the given angle. - The shortest distance from a point to a line is perpendicular. - What angle is formed when the minimum-length side is perpendicular to the other side?

1. The minimum possible value of \(a\) is the perpendicular distance from \(B\) to the ray containing side \(b\). This altitude is opposite \(\alpha\) in a right triangle with hypotenuse \(c\). 2. Therefore, \(a_{\min}=12.0\sin(40^\circ)\approx 7.71\,\text{cm}\). 3. At this minimum, side \(a\) is perpendicular to side \(b\), so the triangle is right and \(\gamma=90^\circ\).

1. The minimum length is approximately \(7.71\,\text{cm}\). 2. The triangle is a right triangle, and \(\gamma=90^\circ\).

- Use the Law of Sines with the given angle and the two known sides. - Can the same sine value belong to more than one angle in a triangle? - Check whether a candidate leaves room for a valid third angle before finding another side. - Keep intermediate angle values unrounded.

1. Let \(a=80\,\text{m}\), \(b=120\,\text{m}\), and \(\alpha=35^\circ\). 2. The Law of Sines gives \(\sin(\beta)=\frac{120\sin(35^\circ)}{80}\approx0.86036\). 3. Both \(\beta_1\approx59.36^\circ\) and \(\beta_2\approx120.64^\circ\) fit with the given angle, so there are two possible triangles. 4. Their third angles are \(\gamma_1\approx85.64^\circ\) and \(\gamma_2\approx24.36^\circ\). 5. Using unrounded angle values gives \(c_1=\frac{80\sin(\gamma_1)}{\sin(35^\circ)}\approx139.07\,\text{m}\) and \(c_2\approx57.52\,\text{m}\).

Two triangles are possible because the SSA data allow two valid angles opposite the longer side. The third side can be \(139.07\,\text{m}\) or \(57.52\,\text{m}\), to the nearest hundredth.

- Express the same altitude from two right triangles in each panel. - For obtuse \(A\), use the acute angle along the extension of the base. - Compare the height expressions, then rearrange the ratios. - A different altitude can connect the remaining pair of sides and opposite angles.

1. In panel a), the two right triangles give \(h=b\sin A\) and \(h=a\sin B\). 2. Equating these heights gives \(b\sin A=a\sin B\), so \(\frac{a}{\sin A}=\frac{b}{\sin B}\). 3. In panel b), the acute angle between \(AC\) and the extension of \(AB\) is \(180^\circ-A\). Hence, \(h=b\sin(180^\circ-A)=b\sin A\). 4. The right triangle from \(B\) still gives \(h=a\sin B\), so the same equality holds when \(A\) is obtuse. 5. If \(A=90^\circ\), the right-triangle sine definition gives the equality directly. 6. Repeating the argument with a different altitude yields \(\frac{b}{\sin B}=\frac{c}{\sin C}\). Combine the relations to obtain the full Law of Sines.

a) \(h=b\sin A=a\sin B\), so \(\frac{a}{\sin A}=\frac{b}{\sin B}\). b) \(h=b\sin(180^\circ-A)=b\sin A=a\sin B\), so the same equality holds. c) A second altitude gives \(\frac{b}{\sin B}=\frac{c}{\sin C}\), hence \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\).

- In \(\triangle BCD\), use the Law of Sines and test the supplementary angle. - In \(\triangle ABD\), all three side lengths are known. - Does determining each triangle also determine which side of \(BD\) it occupies?

1. In \(\triangle BCD\), side \(BD = 6\,\text{cm}\) is opposite \(\angle C = 40^\circ\), and \(BC = 4\,\text{cm}\). 2. By the Law of Sines, \(\sin \angle D = \frac{4\sin 40^\circ}{6} \approx 0.4285\). This gives \(\angle D \approx 25.4^\circ\). The supplementary value is impossible because it would make the angle sum exceed \(180^\circ\). Thus, \(\triangle BCD\) is uniquely determined up to reflection. 3. Triangle \(ABD\) has side lengths \(5\,\text{cm}\), \(3\,\text{cm}\), and \(6\,\text{cm}\). These satisfy the triangle inequality, so \(\triangle ABD\) is uniquely determined by SSS, up to reflection. 4. The two triangles can be placed on the same side or on opposite sides of \(\overline{BD}\). Without an additional condition such as convexity, the entire quadrilateral is not uniquely determined.

Triangle \(BCD\) is uniquely determined by resolving the SSA case with the Law of Sines, and triangle \(ABD\) is uniquely determined by SSS. The quadrilateral is not unique because the component triangles can be placed on the same side or on opposite sides of \(BD\).