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11111nata11111 [884]
3 years ago
12

I have two Law Of Sines Problems that I need to Help Solved! Please!!

Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
6 0

Answer:

From figure A

The value of ∠ B = 75.74°   ,  ∠ C = 70.26° and  AB = 27.37

From figure B

The value of ∠ A = 42.8°   ,  ∠ B = 106.2° and  AC = 30.04

Step-by-step explanation:

<u>Given first figure as :</u>

AC = 28.2

BC = 16.5

∠ A = 34°

Let AB = c

<u>From law of sines</u>

\dfrac{a}{Sin A} = \dfrac{b}{Sin B} = \dfrac{c}{Sin C}

Or, \dfrac{a}{Sin A} = \dfrac{b}{Sin B}

or, \dfrac{16.5}{Sin 34} = \dfrac{28.2}{Sin B}

Or,  29.506 =  \dfrac{28.2}{Sin B}

Or, Sin B =  \dfrac{28.2}{29.5}  

Or, Sin B = 0.955

∴  ∠B = Sin^{-1} 0.955

I.e∠ B = 75.74

Now, ∠ C = 180° - ( ∠A + ∠B )

Or, ∠ C = 180° - ( 34° + 75.74° )

Or, ∠ C = 70.26°

Now, Again

\dfrac{b}{Sin B} = \dfrac{c}{Sin C}

so,  \dfrac{28.2}{Sin 75.74} = \dfrac{c}{Sin 70.26}

Or,   \dfrac{28.2}{0.9691} = \dfrac{c}{0.9412}

Or, c = 29.09 × 0.9412

∴    c = 27.37

I.e AB = 27.37

Hence,  The value of ∠ B = 75.74°   ,  ∠ C = 70.26° and  AB = 27.37

<u>From figure second</u>

Given as :

AB = c= 12

BC = a = 16

∠ C = 31°

let AC = b

<u>From law of sines</u>

\dfrac{a}{Sin A} = \dfrac{b}{Sin B} = \dfrac{c}{Sin C}

Or, \dfrac{a}{SinA } = \dfrac{c}{Sin C}

or,  \dfrac{16}{Sin A} = \dfrac{12}{Sin 31}

or,  \dfrac{16}{Sin A} = \dfrac{12}{0.51}

Or, \dfrac{16}{Sin A} = 23.52

∴ Sin A = \dfrac{16}{23.52}

I.e Sin A = 0.68

Or,  ∠ A = Sin^{-1} 0.68

or,  ∠ A = 42.8°

Now,  ∠ B = 180° - ( 31° + 42.8° )

Or,  ∠ B = 106.2°

Now,  \dfrac{b}{Sin B} = \dfrac{c}{Sin C}

or,  \dfrac{b}{Sin 106.2} = \dfrac{16}{Sin 31}

Or, \dfrac{b}{0.96} = \dfrac{16}{0.51}

or, b = 31.3×0.96

∴ b = 30.04

Hence The value of ∠ A = 42.8°   ,  ∠ B = 106.2° and  AC = 30.04

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