Question

A sailing boat with triangular masts is shown below. Two right triangles can be observed. Triangles ABC and ABD, both right angled at B. The distance BC= 1m, BD = 2m and height AB = 4m 1) The value of Sec D is * 1 point 1/(√5) (√5)/2 √5 2/(√5) 2) The value of cosec C is * 1 point 4/(√17) 1/(√17) √17 (√17)/4 3) The value of Sin² C + cos² D is : * 1 point 0 1 97/85 85/97 4) If Sin P = 3/4 , then Cos P is * 1 point 4/3 7/3 (√7)/4 (√7)/3 5) If 3 TanP = 4, then the value of (3Sinp+2CosP)/(3SinP-2Cos P) * 1 point 4 11/15 7/15 3

Answer 1

SOLUTION

TO DETERMINE

1. If sin P = 3/4 , then cos P is

2. If 3 tanP = 4, then the value of (3 sin P + 2 CosP)/(3SinP-2Cos P)

EVALUATION

1. Here it is given that

sin P = 3/4

Thus we get

cos P

$\displaystyle \sf{ = \sqrt{1 - { \sin}^{2} P} }$

$\displaystyle \sf{ = \sqrt{1 - \frac{9}{16} } }$

$\displaystyle \sf{ = \sqrt{ \frac{16 - 9}{16} } }$

$\displaystyle \sf{ = \sqrt{ \frac{7}{16} } }$

$\displaystyle \sf{ = \frac{ \sqrt{7} }{4} }$

2. Here it is given that 3 tanP = 4

Thus we have

$\displaystyle \sf{ \tan P = \frac{4}{3} }$

Now

$\displaystyle \sf{ \frac{3 \sin P + 2 \cos P}{3 \sin P - 2 \cos P} }$

$\displaystyle \sf{ = \frac{3 \tan P + 2 }{3 \tan P - 2 } }$

$\displaystyle \sf{ = \frac{4 + 2 }{4 - 2 } }$

$\displaystyle \sf{ = \frac{6 }{2 } }$

$\displaystyle \sf{ = 3 }$

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