This is really just about correctly using the buttons on your calculator. For the last three parts when you have the sine of an angle, you need the inverse sine of that number to get the angle. For example,

part d) \(sin\quad p\quad =\quad 0.7\\ so\quad p\quad =\quad { sin }^{ -1 }(0.7)\quad =\quad 44.4\\ \\ \)

a) \(sin\quad 30\quad =\quad \frac { x }{ 10 } \\ so\quad x\quad =\quad 10\quad \times \quad sin\quad 30\quad =\quad 5.00\)

b) \(Cos\quad 40\quad =\quad \frac { 5 }{ x } \\ so\quad x\quad =\quad \frac { 5 }{ cos\quad 40 } =\quad 6.53\)

c) \(Tan\quad x\quad =\quad \frac { 7 }{ 14 } \\ so\quad x\quad =\quad { tan }^{ -1 }\left( \frac { 7 }{ 14 } \right) =\quad 26.6\)

This question is really about encouraging the making of statements that are true about a given situation. In practice you can then pick which ones will be mist hepful to you. Clearly it depends on remmebering the trig ratios - SOHCAHTOA and identifying the Hypotenuse as the longest side and the positions of the Opposite and adject sides relative to the angle in question.

a, \(Sin40\quad =\quad \frac { a }{ 10 } \\ so\quad a\quad =\quad 10\times sin40\\ a\quad =\quad 6.43\quad (3sf)\)

b, \(Cos40\quad =\quad \frac { b }{ 10 } \\ so\quad b\quad =\quad 10\times cos40\\ a\quad =\quad 7.66\quad (3sf)\)

a, \(sin28\quad =\quad \frac { 15 }{ a } \\ so\quad a\quad =\quad \frac { 15 }{ sin28 } \\ a\quad =\quad 32.0\quad (3sf)\)

b, \(tan28\quad =\quad \frac { b }{ 15 } \\ so\quad b\quad =\quad 15\times tan28\\ a\quad =\quad 7.98\quad (3sf)\)

A, \(cos\quad A\quad =\quad \frac { 41 }{ 56 } \\ so,\quad A\quad =\quad { cos }^{ -1 }\left( \frac { 41 }{ 56 } \right) \\ A\quad =\quad 42.9\quad (3sf)\)

B, \(sin\quad B\quad =\quad \frac { 41 }{ 56 } \\ so,\quad B\quad =\quad { sin }^{ -1 }\left( \frac { 41 }{ 56 } \right) \\ A\quad =\quad 47.1\quad (3sf)\)

NOTE, A + B = 90 as they should!

a) \(tan\quad 65\quad =\quad \frac { 5 }{ x } \\ so,\quad x\quad =\quad \frac { 5 }{ tan\quad 65 } \\ A\quad =\quad 2.33\quad (3sf)\)

b) \(sin\quad 65\quad =\quad \frac { 5 }{ l } \\ so,\quad l\quad =\quad \frac { 5 }{ sin\quad 65 } \\ A\quad =\quad 5.52\quad (3sf)\\ \\ OR\quad { l }^{ 2 }\quad =\quad { 5 }^{ 2 }+{ x }^{ 2 }\\ Using\quad the\quad unrounded\quad answer\quad for\quad x\quad gives\quad the\quad same\quad result\quad for\quad l\)

NOTE, you could have used cosine with x and 65°. Again if you used the unrounded anser for x you would have got the same result.

c) \(sin\quad A\quad =\quad \frac { 5 }{ 6 } \\ so,\quad A\quad =\quad { sin }^{ -1 }\left( \frac { 5 }{ 6 } \right) \\ A\quad =\quad 56.4\quad (3sf)\)

d) = \(\frac { 65\quad -\quad 56.4 }{ 65 } \times \quad 100\quad =\quad 13.2\)

a) \({ AC }^{ 2 }\quad =\quad { 4 }^{ 2 }+{ 5 }^{ 2 }\\ AC\quad =\sqrt { 25 } =5.00\quad (3sf)\)

b) \(tan\hat { CAD } =\frac { 4 }{ 3 } \\ so,\quad \hat { CAD } \quad =\quad { tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) \\ \hat { CAD } \quad =\quad 53.1°\)

c) \(tan\hat { mAD } =\frac { 2 }{ 3 } \\ so,\quad \hat { mAD } \quad =\quad { tan }^{ -1 }\left( \frac { 2 }{ 3 } \right) \\ \hat { mAD } \quad =\quad 45.0°\)

d) \(cos22.5\quad =\quad \frac { 3 }{ mA } \\ so,\quad mA\quad =\quad \frac { 3 }{ cos\quad 22.5 } \\ mA\quad =\quad 3.25m\quad (3sf)\)

a) \(tan60\quad =\quad \frac { h }{ 12 } \\ so,\quad h\quad =\quad 12\times tan60\\ h\quad =\quad 20.9m\quad (3sf)\)

b) \(tanD\quad =\quad \frac { 35 }{ 12 } \\ so,\quad D\quad =\quad { tan }^{ -1 }\left( \frac { 35 }{ 12 } \right) \\ h\quad =\quad 71.1°\quad (3sf)\)

a) \(tan40\quad =\quad \frac { 10 }{ x } \\ x\quad =\quad \frac { 10 }{ tan40 } \\ x\quad =\quad 11.9\quad (3sf)\)

\(tan65\quad =\quad \frac { 18 }{ y } \\ y\quad =\quad \frac { 18 }{ tan65 } \\ y\quad =\quad 8.39\quad (3sf)\)

Total distance = 8.39 + 11.9 = 20.29 , 20.3 (3sf)

b) \(tan75\quad =\quad \frac { h+10 }{ 11.9 } \\ h\quad +\quad 10\quad =\quad 11.9\times tan75\\ h\quad =\quad 11.9\times tan75\quad -\quad 10\\ h\quad =\quad 34.4\quad m\quad (3sf)\)

c) \(tanp\quad =\quad \frac { 10\quad +\quad 34.4\quad -\quad 18 }{ 11.9 } =\frac { 26.4 }{ 20.3 } \\ p\quad =\quad { tan }^{ -1 }\left( \frac { 26.4 }{ 20.3 } \right) \\ p\quad =\quad 52.4\quad (3sf)\)