[b][color=#ff0000]28.4 mm[br][br][/color][/b]The 12 mm side is [color=#0000ff][b]O[/b][/color]pposite the 25[sup]o[/sup] angle. The x is the [color=#0000ff][b]H[/b][/color]ypotenuse.[br][br]We know S[b][color=#0000ff]OH[/color][/b] CAH TOA. So we must use Sin as this uses the O and the H.[br][br][math]Sin\theta=\frac{Opp}{Hyp}[/math] so [math]Hyp=\frac{Opp}{Sin\theta}[/math][br][br]So [math]x=\frac{12}{sin25}=28.4mm[/math]

[b][color=#ff0000]18.4 cm[br][br][/color][/b]The 32 cm side is the [b]H[/b]ypotenuse. x is opposite the 35[sup]o[/sup] angle.[br][br]We know S[b][color=#0000ff]OH[/color][/b] CAH TOA. So we must use Sin as this uses the O and the H.[br][br][math]Sin\theta=\frac{Opp}{Hyp}[/math] so [math]Opp=Hyp\times Sin\theta[/math][br][br]So [math]x=32sin35=18.4cm[/math][br]

[color=#ff0000][b]29.8 cm[br][br][/b][/color]The 25 cm side is [b][color=#0000ff]O[/color][/b]pposite the 40[sup]o[/sup] angle. x is [color=#0000ff][b]A[/b][/color]djacent to the 40[sup]o[/sup] angle.[br][br]We know SOH CAH T[color=#0000ff][b]OA[/b][/color]. So we must use Tan as this uses the O and the A.[br][br][math]Tan\theta=\frac{Opp}{Adj}[/math] so [math]Adj=\frac{Opp}{tan\theta}[/math][br][br]So [math]x=\frac{25}{tan40}=29.8cm[/math]

[b][color=#ff0000]6.6 cm[/color][br][br][/b]The 8 cm side is the [b][color=#0000ff]H[/color][/b]ypotenuse. x is [color=#0000ff][b]A[/b][/color]djacent to the 35[sup]o[/sup] angle.[br][br]We know SOH C[color=#0000ff][b]AH[/b][/color] TOA. So we must use Cos as this uses the H and the A.[br][br][math]Cos\theta=\frac{Adj}{Hyp}[/math] so [math]Adj=Hyp\times Cos\theta[/math][br][br]So [math]x=8sin35=6.6cm[/math]

[color=#ff0000][b]5.3 cm[/b][/color][br][br]Remember, Capital letters are for angles, lowercase letters are for sides, and a side has the same letter as the angle which is opposite it.[br]Drawing a sketch can be helpful.[br][br][img]https://image.ibb.co/mOeY1a/Q5.jpg[/img][br]From the sketch it's easy to see the 6 cm side is the [b][color=#0000ff]H[/color][/b]ypotenuse. c is [color=#0000ff][b]A[/b][/color]djacent to the 28[sup]o[/sup] angle.[br][br]We know SOH C[color=#0000ff][b]AH[/b][/color] TOA. So we must use Cos as this uses the H and the A.[br][br][math]Cos\theta=\frac{Adj}{Hyp}[/math] so [math]Adj=Hyp\times Cos\theta[/math][br][br]So [math]c=6sin28=5.3cm[/math]

[color=#ff0000][b]26.6 cm[/b][/color][br][br]Remember, Capital letters are for angles, lowercase letters are for sides, and a side has the same letter as the angle which is opposite it.[br]Drawing a sketch can be helpful.[br][br][img]https://image.ibb.co/bF51nF/Q6.jpg[/img][br][br]From the sketch it's easy to see the 25 cm side is [b][color=#0000ff]O[/color][/b]pposite the 70[sup]o[/sup] angle. q is the [b][color=#0000ff]H[/color][/b]ypotenuse.[br][br]We know S[color=#0000ff][b]OH[/b][/color] CAH TOA. So we must use Sin as this uses the O and the H.[br][br][math]Sin\theta=\frac{Opp}{Hyp}[/math] so [math]Hyp=\frac{Opp}{Sin\theta}[/math][br][br]So [math]q=\frac{25}{sin70}=26.6cm[/math]

[color=#ff0000][b]36.3[sup]o[/sup][/b][br][/color][br]The 16 cm side [color=#0000ff][b]O[/b][/color]pposite angle [math]\theta[/math] and 27 cm is the [b][color=#0000ff]H[/color][/b]ypotenuse.[br]We know S[color=#0000ff][b]OH[/b][/color] CAH TOA. So we must use Sin as this uses the O and the H.[br][math]sin\theta=\frac{Opp}{Hyp}[/math][br][br]so [math]sin\theta=\frac{16}{27}[/math][br][br]so [math]\theta=sin^{-1}\frac{16}{27}=36.3^o[/math]

[color=#ff0000][b]51.3[sup]o[/sup][/b][/color][br][br]The 18 cm side is the [b][color=#0000ff]H[/color][/b]ypotenuse. The 10 cm side is [color=#0000ff][b]A[/b][/color]djacent to angle [math]\theta[/math].[br][br]We know SOH C[color=#0000ff][b]AH[/b][/color] TOA. So we must use Cos as this uses the A and the H.[br][br][math]cos\theta=\frac{adj}{hyp}[/math][br][br]so [math]cos\theta=\frac{10}{16}[/math][br][br]so [math]\theta=cos^{-1}\frac{10}{16}=51.3^o[/math][br]

[color=#ff0000][b]38.0[sup]o[/sup][/b][/color][br][br]The 13 cm is the Hypotenuse. The 8 cm is Opposite angle [math]\theta[/math].[br][br]We know S[color=#0000ff][b]OH[/b][/color] CAH TOA. So we must use Sin as this uses the O and the H.[br][br][math]sin\theta=\frac{Opp}{Hyp}[/math][br][br]So, [math]sin\theta=\frac{8}{13}[/math][br][br]So, [math]\theta=sin^{-1}\frac{8}{13}=38.0[/math]

[color=#ff0000][b]60[sup]o[/sup][/b][/color][br][br]Remember, Capital letters are for angles, lowercase letters are for sides, and a side has the same letter as the angle which is opposite it.[br]Drawing a sketch can be helpful.[br][br][img]https://image.ibb.co/e7Ng4v/Q10.jpg[/img][br]From the diagram it is clear to see 140 mm is the [b][color=#0000ff]H[/color][/b]ypotenuse. 70 mm is [b][color=#0000ff]A[/color][/b]djacent to angle [math]\theta[/math].[br][br]We know SOH C[b][color=#0000ff]AH[/color][/b] TOA. So we must use Cos as this uses the A and the H.[br][br][math]cos\theta=\frac{adj}{hyp}[/math][br][br]So, [math]cos\theta=\frac{70}{140}[/math][br][br]So, [math]\theta=cos^{-1}\frac{70}{140}=60^o[/math][br]