Question

find each measurement indicated. round your answers to the nearest tenth.

1. find ab
2. find ab
3. find ac
4. find m∠c
5. find m∠c
6. find m∠a

solve each triangle. round your answers to the nearest tenth.
16)
17)
18)

Explanation:

Step1: 10번 문제 - 코사인 법칙 적용

$AB^{2}=AC^{2}+BC^{2}-2\cdot AC\cdot BC\cdot\cos C$
$AB^{2}=27^{2}+19^{2}-2\times27\times19\times\cos80^{\circ}$
$AB^{2}=729 + 361-1026\times0.173648$
$AB^{2}=1090 - 178.163848$
$AB^{2}=911.836152$
$AB=\sqrt{911.836152}\approx30.2$ m

Step2: 11번 문제 - 코사인 법칙 적용

$AB^{2}=AC^{2}+BC^{2}-2\cdot AC\cdot BC\cdot\cos C$
$AB^{2}=13^{2}+20^{2}-2\times13\times20\times\cos93^{\circ}$
$AB^{2}=169 + 400-520\times(-0.052336)$
$AB^{2}=569+27.21472$
$AB^{2}=596.21472$
$AB=\sqrt{596.21472}\approx24.4$ in

Step3: 12번 문제 - 코사인 법칙 적용

$AC^{2}=AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot\cos B$
$AC^{2}=14^{2}+8^{2}-2\times14\times8\times\cos117^{\circ}$
$AC^{2}=196 + 64-224\times(-0.45399)$
$AC^{2}=260 + 101.69376$
$AC^{2}=361.69376$
$AC=\sqrt{361.69376}\approx19.0$ in

Step4: 13번 문제 - 코사인 법칙 적용

$\cos C=\frac{AC^{2}+BC^{2}-AB^{2}}{2\cdot AC\cdot BC}$
$AC^{2}=13.7^{2}+14.5^{2}-2\times13.7\times14.5\times\cos97.5^{\circ}$
$AC^{2}=187.69+210.25 - 397.3\times(-0.130526)$
$AC^{2}=397.94+51.85998$
$AC^{2}=449.79998$
$AC=\sqrt{449.79998}\approx21.2$ in
$\cos C=\frac{13.7^{2}+14.5^{2}-21.2^{2}}{2\times13.7\times14.5}=\frac{187.69 + 210.25-449.44}{397.3}=\frac{397.94 - 449.44}{397.3}=\frac{-51.5}{397.3}\approx - 0.13$
$C=\cos^{-1}(-0.13)\approx97.5^{\circ}$

Step5: 14번 문제 - 코사인 법칙 적용

$\cos C=\frac{AC^{2}+BC^{2}-AB^{2}}{2\cdot AC\cdot BC}$
$\cos C=\frac{6^{2}+9^{2}-14^{2}}{2\times6\times9}=\frac{36 + 81-196}{108}=\frac{117 - 196}{108}=\frac{-79}{108}\approx - 0.7315$
$C=\cos^{-1}(-0.7315)\approx137.0^{\circ}$

Step6: 15번 문제 - 코사인 법칙 적용

$\cos A=\frac{AB^{2}+AC^{2}-BC^{2}}{2\cdot AB\cdot AC}$
$\cos A=\frac{28^{2}+16^{2}-28^{2}}{2\times28\times16}=\frac{784 + 256-784}{896}=\frac{256}{896}\approx0.286$
$A=\cos^{-1}(0.286)\approx73.4^{\circ}$

Step7: 16번 문제 - 코사인 법칙으로 각도 찾기

$\cos A=\frac{AB^{2}+AC^{2}-BC^{2}}{2\cdot AB\cdot AC}=\frac{17.7^{2}+28.7^{2}-25.6^{2}}{2\times17.7\times28.7}=\frac{313.29+823.69 - 655.36}{1015.38}=\frac{1136.98 - 655.36}{1015.38}=\frac{481.62}{1015.38}\approx0.474$
$A=\cos^{-1}(0.474)\approx61.7^{\circ}$
$\cos B=\frac{AB^{2}+BC^{2}-AC^{2}}{2\cdot AB\cdot BC}=\frac{17.7^{2}+25.6^{2}-28.7^{2}}{2\times17.7\times25.6}=\frac{313.29+655.36 - 823.69}{906.24}=\frac{968.65 - 823.69}{906.24}=\frac{144.96}{906.24}\approx0.16$
$B=\cos^{-1}(0.16)\approx80.8^{\circ}$
$C = 180^{\circ}-A - B=180^{\circ}-61.7^{\circ}-80.8^{\circ}=37.5^{\circ}$

Step8: 17번 문제 - 코사인 법칙으로 각도 찾기

$\cos A=\frac{AB^{2}+AC^{2}-BC^{2}}{2\cdot AB\cdot AC}=\frac{27^{2}+21^{2}-27^{2}}{2\times27\times21}=\frac{729+441 - 729}{1134}=\frac{441}{1134}\approx0.39$
$A=\cos^{-1}(0.39)\approx67.1^{\circ}$
$B = 180^{\circ}-A - C=180^{\circ}-67.1^{\circ}-33^{\circ}=79.9^{\circ}$
$BC^{2}=AB^{2}+AC^{2}-2\cdot AB\cdot AC\cdot\cos A$
$BC^{2}=27^{2}+21^{2}-2\times27\times21\times0.39$
$BC^{2}=729+441 - 440.19$
$BC^{2}=729.81$
$BC=\sqrt{729.81}\approx27.0$ yd

Step9: 18번 문제 - 코사인 법칙으로 각도 찾기

$AB^{2}=AC^{2}+BC^{2}-2\cdot AC\cdot BC\cdot\cos C$
$AB^{2}=27^{2}+13^{2}-2\times27\times13\times\cos106^{\circ}$
$AB^{2}=729+169 - 702\times(-0.275637)$
$AB^{2}=898+193.497174$
$AB^{2}=1091.497174$
$AB=\sqrt{1091.497174}\approx33.0$ in
$\cos A=\frac{AB^{2}+AC^{2}-BC^{2}}{2\cdot AB\cdot AC}=\frac{33.0^{2}+27^{2}-13^{2}}{2\times33.0\times27}=\frac{1089+729 - 169}{1782}=\frac{1649}{1782}\approx0.925$
$A=\cos^{-1}(0.925)\approx22.4^{\circ}$
$B = 180^{\circ}-A - C=180^{\circ}-22.4^{\circ}-106^{\circ}=51.6^{\circ}$

Answer:

1. $AB\approx30.2$ m
2. $AB\approx24.4$ in
3. $AC\approx19.0$ in
4. $m\angle C\approx97.5^{\circ}$
5. $m\angle C\approx137.0^{\circ}$
6. $m\angle A\approx73.4^{\circ}$
7. $A\approx61.7^{\circ}$, $B\approx80.8^{\circ}$, $C = 37.5^{\circ}$
8. $A\approx67.1^{\circ}$, $B\approx79.9^{\circ}$, $BC\approx27.0$ yd
9. $AB\approx33.0$ in, $A\approx22.4^{\circ}$, $B\approx51.6^{\circ}$