Question

solve δabc. give lengths to the nearest tenth. give angle measures in degrees and minutes.

1. ( a = 71^circ ), ( a = 20 ), ( c = 62^circ )
2. ( a = 4 ), ( b = 2 ), ( c = 5 )
3. ( a = 12 ), ( b = 110^circ ), ( c = 35^circ )
4. ( b = 115^circ ), ( a = 18^circ ), ( a = 4 )
5. ( b = 7 ), ( c = 10 ), ( a = 51^circ )
6. ( b = 9 ), ( c = 11 ), ( a = 123^circ )

Explanation:

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Problem 1: $A = 71^\circ, a = 20, C = 62^\circ$

Step1: Find angle $B$

Sum of angles in $\triangle ABC$ is $180^\circ$.
$B = 180^\circ - 71^\circ - 62^\circ = 47^\circ$

Step2: Find side $b$ via Law of Sines

$\frac{b}{\sin B} = \frac{a}{\sin A}$
$b = \frac{a \cdot \sin B}{\sin A} = \frac{20 \cdot \sin 47^\circ}{\sin 71^\circ} \approx \frac{20 \cdot 0.7314}{0.9455} \approx 15.5$

Step3: Find side $c$ via Law of Sines

$\frac{c}{\sin C} = \frac{a}{\sin A}$
$c = \frac{a \cdot \sin C}{\sin A} = \frac{20 \cdot \sin 62^\circ}{\sin 71^\circ} \approx \frac{20 \cdot 0.8829}{0.9455} \approx 18.7$

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Problem 2: $a = 4, b = 2, c = 5$

Step1: Find angle $A$ via Law of Cosines

$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
$\cos A = \frac{2^2 + 5^2 - 4^2}{2(2)(5)} = \frac{4 + 25 - 16}{20} = 0.65$
$A = \arccos(0.65) \approx 49^\circ 27'$

Step2: Find angle $B$ via Law of Cosines

$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$
$\cos B = \frac{4^2 + 5^2 - 2^2}{2(4)(5)} = \frac{16 + 25 - 4}{40} = 0.925$
$B = \arccos(0.925) \approx 22^\circ 1'$

Step3: Find angle $C$ via angle sum

$C = 180^\circ - 49^\circ 27' - 22^\circ 1' = 108^\circ 32'$

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Problem 3: $a = 12, B = 110^\circ, C = 35^\circ$

Step1: Find angle $A$

$A = 180^\circ - 110^\circ - 35^\circ = 35^\circ$

Step2: Find side $b$ via Law of Sines

$\frac{b}{\sin B} = \frac{a}{\sin A}$
$b = \frac{12 \cdot \sin 110^\circ}{\sin 35^\circ} \approx \frac{12 \cdot 0.9397}{0.5736} \approx 19.7$

Step3: Find side $c$ via Law of Sines

$\frac{c}{\sin C} = \frac{a}{\sin A}$
$c = \frac{12 \cdot \sin 35^\circ}{\sin 35^\circ} = 12.0$

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Problem 4: $B = 115^\circ, A = 18^\circ, a = 4$

Step1: Find angle $C$

$C = 180^\circ - 115^\circ - 18^\circ = 47^\circ$

Step2: Find side $b$ via Law of Sines

$\frac{b}{\sin B} = \frac{a}{\sin A}$
$b = \frac{4 \cdot \sin 115^\circ}{\sin 18^\circ} \approx \frac{4 \cdot 0.9063}{0.3090} \approx 11.7$

Step3: Find side $c$ via Law of Sines

$\frac{c}{\sin C} = \frac{a}{\sin A}$
$c = \frac{4 \cdot \sin 47^\circ}{\sin 18^\circ} \approx \frac{4 \cdot 0.7314}{0.3090} \approx 9.5$

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Problem 5: $b = 7, c = 10, A = 51^\circ$

Step1: Find side $a$ via Law of Cosines

$a^2 = b^2 + c^2 - 2bc\cos A$
$a^2 = 7^2 + 10^2 - 2(7)(10)\cos 51^\circ \approx 49 + 100 - 140 \cdot 0.6293 = 149 - 88.10 = 60.90$
$a \approx \sqrt{60.90} \approx 7.8$

Step2: Find angle $B$ via Law of Sines

$\sin B = \frac{b \cdot \sin A}{a}$
$\sin B = \frac{7 \cdot \sin 51^\circ}{7.8} \approx \frac{7 \cdot 0.7771}{7.8} \approx 0.7023$
$B \approx 44^\circ 39'$

Step3: Find angle $C$ via angle sum

$C = 180^\circ - 51^\circ - 44^\circ 39' = 84^\circ 21'$

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Problem 6: $b = 9, c = 11, A = 123^\circ$

Step1: Find side $a$ via Law of Cosines

$a^2 = b^2 + c^2 - 2bc\cos A$
$a^2 = 9^2 + 11^2 - 2(9)(11)\cos 123^\circ \approx 81 + 121 - 198 \cdot (-0.5446) = 202 + 107.83 = 309.83$
$a \approx \sqrt{309.83} \approx 17.6$

Step2: Find angle $B$ via Law of Sines

$\sin B = \frac{b \cdot \sin A}{a}$
$\sin B = \frac{9 \cdot \sin 123^\circ}{17.6} \approx \frac{9 \cdot 0.8387}{17.6} \approx 0.4270$
$B \approx 25^\circ 17'$

Step3: Find angle $C$ via angle sum

$C = 180^\circ - 123^\circ - 25^\circ 17' = 31^\circ 43'$

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Answer:

1. $B = 47^\circ$, $b \approx 15.5$, $c \approx 18.7$
2. $A \approx 49^\circ 27'$, $B \approx 22^\circ 1'$, $C \approx 108^\circ 32'$
3. $A = 35^\circ$, $b \approx 19.7$, $c = 12.0$
4. $C = 47^\circ$, $b \approx 11.7$, $c \approx 9.5$
5. $a \approx 7.8$, $B \approx 44^\circ 39'$, $C \approx 84^\circ 21'$
6. $a \approx 17.6$, $B \approx 25^\circ 17'$, $C \approx 31^\circ 43'$