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How would you go about solving this? Please provide formulae, rules, explanation / working out . Thank you in advance!

100 POINTS ENSURE YOUR ANSWER IS CORRECT BEFORE POSTING Past Papers How would you go about solving this Please provide formulae rules explanation working out T class=

Respuesta :

Answer :

  • 28.2°

Explanation :

first , let's find the length of the side 'AC'.

using the law of cosines

  • b = √(a^2 + c^2 -2ac*cos B)

here,

a = 12.3 m

b = AC

c = 9.7 m

A = 115°

plug in,

AC = √((12.3 m)^2 + (9.7m)^2 - 2*12.3m*9.7m*cos(115°)

AC = √(245.38 m^2 - (-100.845 m^2)

  • AC = 18.607 m

now, we can find the measure of angle C or x using the law of sines

  • sinB/b = sinC/c

B = 115°

b = 18.607 m

C = x°

c = 9.7m

plug in,

sin(115°)/18.607m = sin(x)/9.7m

x = sin^-1[(sin(115°)*9.7m)/18.607m)]

  • x = 28.2° ( 3 s.f. )

Answer:

x = 28.2

Step-by-step explanation:

To find the value of x, first find the measure of side AC using the Law of Cosines, then use the Law of Sines to find the measure of angle C.

[tex]\boxed{\begin{array}{l}\underline{\textsf{Law of Cosines}}\\\\b^2=a^2+c^2-2ac \cos B\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$a, b$ and $c$ are the sides.}\\\phantom{ww}\bullet\;\textsf{$B$ is the angle opposite side $b$.}\end{array}}[/tex]

In this case:

[tex]a = BC = 12.3\\\\b = AC\\\\c = AB = 9.7\\\\B = 115^{\circ}[/tex]

Substitute the values into the formula and solve for the exact measure of AC:

[tex]AC^2=12.3^2+9.7^2-2(12.3)(9.7)\cos 115^{\circ}\\\\\\AC^2=151.29+94.09-238.62\cos 115^{\circ}\\\\\\AC^2=245.38-238.62\cos 115^{\circ}\\\\\\AC=\sqrt{245.38-238.62\cos 115^{\circ}}[/tex]

[tex]\boxed{\begin{array}{l}\underline{\textsf{Law of Sines}} \\\\\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{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]

in this case:

[tex]B = 115^{\circ}\\\\b = AC=\sqrt{245.38-238.62\cos 115^{\circ}}\\\\C = x^{\circ}\\\\c = AB=9.7[/tex]

Substitute the values into the formula and solve for x:

[tex]\dfrac{\sin 115^{\circ}}{\sqrt{245.38-238.62\cos 115^{\circ}}}=\dfrac{\sin x^{\circ}}{9.7}\\\\\\\\\sin x^{\circ}=\dfrac{9.7\sin 115^{\circ}}{\sqrt{245.38-238.62\cos 115^{\circ}}}\\\\\\\\x^{\circ}=\sin^{-1}\left(\dfrac{9.7\sin 115^{\circ}}{\sqrt{245.38-238.62\cos 115^{\circ}}}\right)\\\\\\\\x^{\circ}=28.1943172...^{\circ}\\\\\\x^{\circ}=28.2^{\circ}\\\\\\x=28.2\; \sf (3\;s.f.)[/tex]

Therefore, the value of x correct to 3 significant figures is:

[tex]\Large\boxed{\boxed{x=28.2}}[/tex]

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