Question

questions 8 - 10, deal with the ambiguous ssa case. for each, find all possible solutions and sketch the triangle(s) in each case. (3 pbints each)

1. a = 50

, a = 15, b = 12

1. a = 50

, a = 10, b = 15

1. a = 47

, a = 1.6, b = 2

Explanation:

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Question 8

Step1: Use Law of Sines for $\sin B$

$\sin B = \frac{b \sin A}{a} = \frac{12 \sin 50^\circ}{15}$

Step2: Calculate $\sin B$ value

$\sin B \approx \frac{12 \times 0.7660}{15} \approx 0.6128$

Step3: Find $B$ and check validity

$B \approx \arcsin(0.6128) \approx 36.9^\circ$. Since $a > b$, only 1 $B$ exists.

Step4: Calculate angle $C$

$C = 180^\circ - A - B \approx 180^\circ - 50^\circ - 36.9^\circ = 93.1^\circ$

Step5: Use Law of Sines for $c$

$c = \frac{a \sin C}{\sin A} \approx \frac{15 \sin 93.1^\circ}{\sin 50^\circ} \approx \frac{15 \times 0.9985}{0.7660} \approx 19.6$

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Question 9

Step1: Use Law of Sines for $\sin B$

$\sin B = \frac{b \sin A}{a} = \frac{15 \sin 50^\circ}{10}$

Step2: Calculate $\sin B$ value

$\sin B \approx \frac{15 \times 0.7660}{10} \approx 1.149$

Step3: Check validity of $\sin B$

Since $\sin B > 1$, no valid angle $B$ exists. No triangles possible.

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Question 10

Step1: Use Law of Sines for $\sin B$

$\sin B = \frac{b \sin A}{a} = \frac{2 \sin 47^\circ}{1.6}$

Step2: Calculate $\sin B$ value

$\sin B \approx \frac{2 \times 0.7314}{1.6} \approx 0.9142$

Step3: Find two possible $B$ values

$B_1 \approx \arcsin(0.9142) \approx 66.8^\circ$, $B_2 = 180^\circ - 66.8^\circ = 113.2^\circ$

Step4: Calculate $C_1$ and $c_1$

$C_1 = 180^\circ - 47^\circ - 66.8^\circ = 66.2^\circ$
$c_1 = \frac{a \sin C_1}{\sin A} \approx \frac{1.6 \sin 66.2^\circ}{\sin 47^\circ} \approx \frac{1.6 \times 0.9141}{0.7314} \approx 1.99$

Step5: Calculate $C_2$ and $c_2$

$C_2 = 180^\circ - 47^\circ - 113.2^\circ = 19.8^\circ$
$c_2 = \frac{a \sin C_2}{\sin A} \approx \frac{1.6 \sin 19.8^\circ}{\sin 47^\circ} \approx \frac{1.6 \times 0.3380}{0.7314} \approx 0.72$

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Sketch Notes:

• Q8: Triangle with angle $A=50^\circ$, side $a=15$ (longest side so far), side $b=12$, angle $B$ acute.
• Q9: No triangle can be drawn.
• Q10: Two triangles: one with acute $B_1$, one with obtuse $B_2$, both sharing angle $A=47^\circ$, side $a=1.6$, side $b=2$.

Answer:

Question 8:

• $B \approx 36.9^\circ$, $C \approx 93.1^\circ$, $c \approx 19.6$ (one valid triangle)

Question 9:

No valid triangles

Question 10:

• Triangle 1: $B_1 \approx 66.8^\circ$, $C_1 \approx 66.2^\circ$, $c_1 \approx 1.99$
• Triangle 2: $B_2 \approx 113.2^\circ$, $C_2 \approx 19.8^\circ$, $c_2 \approx 0.72$