Question

1. three parts of $\triangle abc$ are given. assume $a$ is opposite $\angle a$, $b$ is opposite $\angle b$, $c$ is opposite $\angle c$.

a) $\angle a = 50^\circ$, $a = 35$, $c = 27$, find $\angle b$.
b) $a = 23$, $b = 31$, $c = 19$, find $\angle a$
c) $\angle b = 94^\circ$, $a = 29$, $c = 21$, find $\angle c$.
d) $\angle c = 21^\circ$, $\angle a = 48^\circ$, $b = 26$, find $c$.
e) $\angle b = 108^\circ$, $\angle c = 34^\circ$, $b = 68$, find $a$.
f) $\angle a = 80^\circ$, $a = 57$, $c = 22$, find $\angle b$
g) $\angle c = 73^\circ$, $a = 12$, $b = 14$, find $c$
h) $a = 10$, $b = 12$, $c = 17$, find $\angle c$

Explanation:

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Part (a)

Step1: Apply Law of Sines

$\frac{\sin C}{c} = \frac{\sin A}{a}$
$\sin C = \frac{c \cdot \sin A}{a} = \frac{27 \cdot \sin 50^\circ}{35}$
$\sin C \approx \frac{27 \cdot 0.7660}{35} \approx 0.5909$
$\angle C \approx \arcsin(0.5909) \approx 36.2^\circ$

Step2: Calculate $\angle B$

$\angle B = 180^\circ - \angle A - \angle C$
$\angle B \approx 180^\circ - 50^\circ - 36.2^\circ = 93.8^\circ \approx 94^\circ$

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Part (b)

Step1: Apply Law of Cosines

$a^2 = b^2 + c^2 - 2bc\cos A$
$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
$\cos A = \frac{31^2 + 19^2 - 23^2}{2 \cdot 31 \cdot 19} = \frac{961 + 361 - 529}{1178} = \frac{793}{1178} \approx 0.6732$

Step2: Find $\angle A$

$\angle A = \arccos(0.6732) \approx 47.7^\circ \approx 48^\circ$

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Part (c)

Step1: Apply Law of Cosines for $b$

$b^2 = a^2 + c^2 - 2ac\cos B$
$b^2 = 29^2 + 21^2 - 2 \cdot 29 \cdot 21 \cdot \cos 94^\circ$
$b^2 = 841 + 441 - 1218 \cdot (-0.0698) \approx 1282 + 85.0 \approx 1367$
$b \approx \sqrt{1367} \approx 36.97$

Step2: Apply Law of Sines for $\angle C$

$\frac{\sin C}{c} = \frac{\sin B}{b}$
$\sin C = \frac{21 \cdot \sin 94^\circ}{36.97} \approx \frac{21 \cdot 0.9976}{36.97} \approx 0.570$
$\angle C \approx \arcsin(0.570) \approx 34.8^\circ \approx 35^\circ$

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Part (d)

Step1: Calculate $\angle B$

$\angle B = 180^\circ - \angle A - \angle C$
$\angle B = 180^\circ - 48^\circ - 21^\circ = 111^\circ$

Step2: Apply Law of Sines for $c$

$\frac{c}{\sin C} = \frac{b}{\sin B}$
$c = \frac{b \cdot \sin C}{\sin B} = \frac{26 \cdot \sin 21^\circ}{\sin 111^\circ}$
$c \approx \frac{26 \cdot 0.3584}{0.9336} \approx 9.9$

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Part (e)

Step1: Calculate $\angle A$

$\angle A = 180^\circ - \angle B - \angle C$
$\angle A = 180^\circ - 108^\circ - 34^\circ = 38^\circ$

Step2: Apply Law of Sines for $a$

$\frac{a}{\sin A} = \frac{b}{\sin B}$
$a = \frac{b \cdot \sin A}{\sin B} = \frac{68 \cdot \sin 38^\circ}{\sin 108^\circ}$
$a \approx \frac{68 \cdot 0.6157}{0.9511} \approx 44.7$

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Part (f)

Step1: Apply Law of Sines for $\angle C$

$\frac{\sin C}{c} = \frac{\sin A}{a}$
$\sin C = \frac{22 \cdot \sin 80^\circ}{57} \approx \frac{22 \cdot 0.9848}{57} \approx 0.379$
$\angle C \approx \arcsin(0.379) \approx 22.3^\circ$

Step2: Calculate $\angle B$

$\angle B = 180^\circ - \angle A - \angle C$
$\angle B \approx 180^\circ - 80^\circ - 22.3^\circ = 77.7^\circ \approx 78^\circ$

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Part (g)

Step1: Apply Law of Cosines for $c$

$c^2 = a^2 + b^2 - 2ab\cos C$
$c^2 = 12^2 + 14^2 - 2 \cdot 12 \cdot 14 \cdot \cos 73^\circ$
$c^2 = 144 + 196 - 336 \cdot 0.2924 \approx 340 - 98.2 \approx 241.8$
$c \approx \sqrt{241.8} \approx 15.55$

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Part (h)

Step1: Apply Law of Cosines

$c^2 = a^2 + b^2 - 2ab\cos C$
$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
$\cos C = \frac{10^2 + 12^2 - 17^2}{2 \cdot 10 \cdot 12} = \frac{100 + 144 - 289}{240} = \frac{-45}{240} = -0.1875$

Step2: Find $\angle C$

$\angle C = \arccos(-0.1875) \approx 100.8^\circ \approx 101^\circ$

Answer:

a) $\angle B \approx 94^\circ$
b) $\angle A \approx 48^\circ$
c) $\angle C \approx 35^\circ$
d) $c \approx 9.9$
e) $a \approx 44.7$
f) $\angle B \approx 78^\circ$
g) $c \approx 15.6$
h) $\angle C \approx 101^\circ$