Answer:
11) A = 10.19°, C = 154.31°, c = 11.03
12) A = 174.68°, C = 2.57°, a = 11.99
13) A = 25.57°, B = 9.43°, a = 10.53
14) B = 75.48°, C = 4.52°, b = 122.87
15) B = 18.22°, C = 51.53°, c = 40.05
Step-by-step explanation:
The Law of Sines states that, in any triangle, the ratio of the length of a side to the sine of its opposite angle is constant across all three sides:
[tex]\boxed{\begin{array}{l}\underline{\textsf{Law of Sines}} \\\\\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\;\;\textsf{or}\;\;\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\\\\\textsf{where:}\\\phantom{ww}\bullet \;\textsf{$A, B$ and $C$ are the angles.}\\\phantom{ww}\bullet\;\textsf{$a, b$ and $c$ are the sides opposite the angles.}\end{array}}[/tex]
To solve a triangle using the Law of Sines, substitute the given values into the appropriate equation.
Question 11
Given:
- B = 15°30' = 15.5°
- a = 4.5
- b = 6.8
Substitute the values into the equation and solve for A:
[tex]\dfrac{\sin A}{4.5}=\dfrac{\sin 15.5^{\circ}}{6.8}\\\\\\A=\sin^{-1}\left(\dfrac{4.5\sin 15.5^{\circ}}{6.8}\right)\\\\\\A=10.186272127686...^{\circ}\\\\\\A=10.19^{\circ}\;\sf (2\;d.p.)[/tex]
Angles in a triangle sum to 180°. Therefore:
[tex]C=180^{\circ}-A-B\\\\C=180^{\circ}-10.19^{\circ}-15.5^{\circ}\\\\C=154.31^{\circ}[/tex]
Use the Law of Sines again to find the value of c:
[tex]\dfrac{6.8}{\sin 15.5^{\circ}}=\dfrac{c}{\sin 154.31^{\circ}}\\\\\\c=\dfrac{6.8\sin 154.31^{\circ}}{\sin 15.5^{\circ}}\\\\\\c=11.03\; \sf (2\;d.p.)[/tex]
[tex]\hrulefill[/tex]
Question 12
Given:
- B = 2°45' = 2.75°
- b = 6.2
- c = 5.8
Substitute the values into the equation and solve for C:
[tex]\dfrac{\sin 2.75^{\circ}}{6.2}=\dfrac{\sin C}{5.8}\\\\\\C=\sin^{-1}\left(\dfrac{5.8\sin 2.75^{\circ}}{6.2}\right)\\\\\\C=2.57245720973...^{\circ}\\\\\\C=2.57^{\circ}\;\sf (2\;d.p.)[/tex]
Angles in a triangle sum to 180°. Therefore:
[tex]A=180^{\circ}-B-C\\\\A=180^{\circ}-2.75^{\circ}-2.5724572...^{\circ}\\\\A=174.67754279...^{\circ}\\\\A=174.68^{\circ}\;\sf (2\;d.p.)[/tex]
Use the Law of Sines again to find the value of a:
[tex]\dfrac{a}{\sin (174.67754279...^{\circ})}=\dfrac{6.2}{\sin 2.75^{\circ}}\\\\\\a=\dfrac{6.2\sin (174.67754279...^{\circ})}{\sin 2.75^{\circ}}\\\\\\a=11.987072659...\\\\\\a=11.99\; \sf (2\;d.p.)[/tex]
[tex]\hrulefill[/tex]
Question 13
Given:
Substitute the values into the equation and solve for B:
[tex]\dfrac{\sin B}{4}=\dfrac{\sin 145^{\circ}}{14}\\\\\\B=\sin^{-1}\left(\dfrac{4\sin 145^{\circ}}{14}\right)\\\\\\B=9.4321184407...^{\circ}\\\\\\B=9.43^{\circ}\;\sf (2\;d.p.)[/tex]
Angles in a triangle sum to 180°. Therefore:
[tex]A=180^{\circ}-B-C\\\\A=180^{\circ}-9.4321184407...^{\circ}-145^{\circ}\\\\A=25.56788155927...^{\circ}\\\\A=25.57^{\circ}\;\sf (2\;d.p.)[/tex]
Use the Law of Sines again to find the value of a:
[tex]\dfrac{a}{\sin (25.56788155927...^{\circ})}=\dfrac{14}{\sin 145^{\circ}}\\\\\\a=\dfrac{14\sin (25.56788155927...^{\circ})}{\sin 145^{\circ}}\\\\\\a=10.53411813...\\\\\\a=10.53\; \sf (2\;d.p.)[/tex]
[tex]\hrulefill[/tex]
Question 14
Given:
Substitute the values into the equation and solve for C:
[tex]\dfrac{\sin 100^{\circ}}{125}=\dfrac{\sin C}{10}\\\\\\C=\sin^{-1}\left(\dfrac{10 \sin 100^{\circ}}{125}\right)\\\\\\C=4.5187090946...^{\circ}\\\\\\C=4.52^{\circ}\;\sf (2\;d.p.)[/tex]
Angles in a triangle sum to 180°. Therefore:
[tex]B=180^{\circ}-A-C\\\\B=180^{\circ}-100^{\circ}-4.5187090946...^{\circ}\\\\B=75.48129090538...^{\circ}\\\\B=75.48^{\circ}\;\sf (2\;d.p.)[/tex]
Use the Law of Sines again to find the value of b:
[tex]\dfrac{b}{\sin (75.48129090538...^{\circ})}=\dfrac{125}{\sin 100^{\circ}}\\\\\\b=\dfrac{125\sin (75.48129090538...^{\circ})}{\sin 100^{\circ}}\\\\\\b=122.874975838..\\\\\\b=122.87\; \sf (2\;d.p.)[/tex]
[tex]\hrulefill[/tex]
Question 15
Given:
- A = 110°15' = 110.25°
- a = 48
- b = 16
Substitute the values into the equation and solve for B:
[tex]\dfrac{\sin 110.25^{\circ}}{48}=\dfrac{\sin B}{16}\\\\\\B=\sin^{-1}\left(\dfrac{16 \sin 110.25^{\circ}}{48}\right)\\\\\\B=18.223857089...^{\circ}\\\\\\B=18.22^{\circ}\;\sf (2\;d.p.)[/tex]
Angles in a triangle sum to 180°. Therefore:
[tex]C=180^{\circ}-A-B\\\\C=180^{\circ}-110.25^{\circ}-18.223857089...^{\circ}\\\\C=51.5261429107...^{\circ}\\\\C=51.53^{\circ}\;\sf (2\;d.p.)[/tex]
Use the Law of Sines again to find the value of c:
[tex]\dfrac{c}{\sin (51.5261429107...^{\circ})}=\dfrac{48}{\sin 110.25^{\circ}}\\\\\\c=\dfrac{48\sin (51.5261429107...^{\circ})}{\sin 110.25^{\circ}}\\\\\\c=40.054539135...\\\\\\c=40.05\; \sf (2\;d.p.)[/tex]
