Question

law of sines and law of cosines

Answer 1

Answer:

Step-by-step explanation:

$1.\\ \\ \frac{sin148}{25}=\frac{sinx}{12}\\ \\ x=sin^{-1}\frac{12sin148}{12}\\ \\ x=32^o\\ \\ 2.\\ \\ x^2=42^2+16^2-2(42)16cos120\\ \\ x^2=2692\\ \\ x=\sqrt{2692}\\ \\ 3.\\ \\ 19^2=17^2+14^2-2(17)14cosx\\ \\ x=cos^{-1}\frac{124}{476}\\ \\ x=74.9^o\\ \\ 4.\\ \\ \frac{sin22}{15}=\frac{sin(180-48-22}{x}\\ \\ x=\frac{15sin110}{sin22}\\ \\ x\approx 37.63$

Answer 2

Answer:

4. $\displaystyle 38 ≈ x$

3. $\displaystyle 75° ≈ m∠X$

2. $\displaystyle 52 ≈ x$

1. $\displaystyle 15° ≈ m∠X$

Step-by-step explanation:

Before we begin, here are the formulas for both laws:

Solving for Angles

$\displaystyle \frac{x^2 + y^2 - z^2}{2xy} = cos∠Z \\ \frac{x^2 - y^2 + z^2}{2xz} = cos∠Y \\ \frac{-x^2 + y^2 + z^2}{2yz} = cos∠X$

Do not forget to use $\displaystyle cos^{-1}$in the end or you will throw the result off!

Solving for Edges

$\displaystyle y^2 + x^2 - 2yx\:cos∠Z = z^2 \\ z^2 + x^2 - 2zx\:cos∠Y = y^2\\ z^2 + y^2 - 2zy\:cos∠X = x^2$

Take the square root of the end result or you will throw it off!

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Solving for Angles

$\displaystyle \frac{sin∠Z}{z} = \frac{sin∠Y}{y} = \frac{sin∠X}{x}$

Do not forget to use $\displaystyle sin^{-1}$in the end or you will throw the result off!

Solving for Edges

$\displaystyle \frac{z}{sin∠Z} = \frac{y}{sin∠Y} = \frac{x}{sin∠X}$

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4. $\displaystyle \frac{15}{sin\:22} = \frac{x}{sin\:110} \hookrightarrow \frac{15sin\:110}{sin\:22} = x \hookrightarrow 37,627178911... = x \\ \\ 38 ≈ x$

3. $\displaystyle \frac{-19^2 + 17^2 + 14^2}{2[17][14]} = cos∠X \hookrightarrow \frac{-361 + 289 + 196}{476} = cos∠X \hookrightarrow 0,2605042016... = cos∠X; 74,900018217...° = cos^{-1}\:0,2605042016... \\ \\ 75° ≈ m∠X$

2. $\displaystyle 16^2 + 42^2 - 2[42][16]\:cos\:120 = x^2 \hookrightarrow 256 + 1764 - 1344\:cos\:120 = x^2 \hookrightarrow \sqrt{2692} = \sqrt{x^2}; 2\sqrt{673}\:[or\:51,884487084...] = x \\ \\ 52 ≈ x$

You could have also used the Law of Sines sinse the base angles are congruent [30°].

1. $\displaystyle \frac{sin\:148}{25} = \frac{sin∠X}{12} \hookrightarrow \frac{12sin\:148}{25} = sin∠X \hookrightarrow 0,2543612468... = sin∠X; 14,735739332...° = sin^{-1}\:0,2543612468... \\ \\ 15° ≈ m∠X$

I am joyous to assist you at any time.