Can anyone please help with these 3 unit trig questions (1 Viewer)

dabatman · May 13, 2017

I have attached the questions, can anyone please help?

Thanks in advance

ProdigyInspired · May 13, 2017

For 13a), use equal side and interior angle rule on ADP.

Mahan1 · May 14, 2017

$\textrm{Q22. If you draw what's happening this is what you get:}$

$\textrm{Then observe triangles BCQ , ABC and APB are isosceles,}$

$AP = AB =1, BC = CQ \ \textrm{and} \ BC= AB =1.$

$\textrm{Also APQ, ABC and PBQ are right angle triangles. }$

$\textrm{since the angle CAB is 45 then} AB = BC =1, \textrm{hence}\ AC = \sqrt{2}$

$\textrm{similarly in triangle BCQ,} \ CQ = BC =1. \textrm{therefore:}$

$AP =1, AQ = 1+ \sqrt{2} \Rightarrow PQ =\sqrt{1^{2} + (1+\sqrt{2})^{2}} = 2.61km \ \textrm{(correct to 2 decimal places)}$

$\textrm{Q13)b Note ABCD is a parallelogram so the adjacent angles are co-interior hence:}$

$\textrm{angle} \ DCB =120, BC=3 \ \textrm{and} \ PC=6$

$\textrm{Apply cosine rule to triangle APB}:$

$AB^{2} = AP^{2} + PB^{2} - 2 \ AP \ PB \ cos x, \textrm{then solve for cosx}$

$\textrm{20a) Using sine rule in triangle ABD}$

$\frac{sin(\alpha - \beta )}{x} = \frac{sin\beta}{BD} \ \textrm{Since} \ BD = \frac{h}{sin\alpha} \Rightarrow \frac{sin(\alpha - \beta )}{x} = \frac{sin\beta}{\frac{h}{sin\alpha}}$

$\textrm{b) is just a substitution and for c) just equate a) and b) }$

$\textrm{for the last question, part a) use at P and Q the North direct are parallel}$

$\textrm{and form a co-interior angles and the angle of revolution at point Q is 360}$

$\textrm{Note the angle LPQ is 20 then use sine rule}$

$\textrm{For part c) note that at that time that forms a right angled triangle but the angle at point P is still 20}$

$\textrm{and you already know what PL is from part b) then } tan(\theta) *PL = \ \textrm{Distance from the Light house}$

Can anyone please help with these 3 unit trig questions (1 Viewer)

dabatman

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ProdigyInspired

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Mahan1

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