MATHEMATICS HIGH SCHOOL

6) Two coastguard stations P and Q are 17km apart, with due East of P. A ship S is observed in distress on a bearing 048° fromP and 324 from Q. How far is the ship from each of the coastguard stations?

Answers

Answer 1

Answer:

Answer:

The ship S is at 10.05 km to coastguard P, and 12.70 km to coastguard Q.

Step-by-step explanation:

Let the distance of the ship to coastguard P be represented by x, and its distance to coastguard Q be represented by y.

But,

<P = 048°

<Q = $360^(o)$ - $324^(o)$

= 0 $36^(o)$

Sum of angles in a triangle = $180^(o)$

<P + <Q + <S = $180^(o)$

048° + 0 $36^(o)$ + <S = $180^(o)$

$84^(o)$ + <S = $180^(o)$

<S = $180^(o)$ - $84^(o)$

= $96^(o)$

<S = $96^(o)$

Applying the Sine rule,

$(y)/(Sin P)$ = $(x)/(Sin Q)$ = $(z)/(Sin S)$

$(y)/(Sin P)$ = $(z)/(Sin S)$

$(y)/(Sin 48^(o) )$ = $(17)/(Sin 96^(o) )$

$(y)/(0.74314)$ = $(17)/(0.99452)$

⇒ y = $(12.63338)/(0.99452)$

= 12.703

y = 12.70 km

$(x)/(Sin Q)$ = $(z)/(Sin S)$

$(x)/(Sin 36^(o) )$ = $(17)/(Sin 96^(o) )$

$(x)/(0.58779)$ = $(17)/(0.99452)$

⇒ x = $(9.992430)/(0.99452)$

= 10.0475

x = 10.05 km

Thus,

the ship S is at a distance of 10.05 km to coastguard P, and 12.70 km to coastguard Q.

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6.3=0.9y
y=

Answers

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Which three measurements could be the lengths of the sides of a triangle? A.5cm,3cm,9cm
B.5cm,2cm,7cm
C.5cm,2cm,2cm
D.5cm,3cm,7cm

Answers

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B). Technically yes, but not really. End-to-end, the 5 and the 2 would lay down
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D). Yes. A cute little scalene triangle.

$a,b,c-lengts\ of\ the\ sides\ of\ a\ triangle\n\nIf\ a\leq b\leq c\ then\ a+b > c\n\nA.\n5+3=8\not>9-NO\nB.\n5+2=7\not>7-NO\nC.\n2+2=4\not>5-NO\nD.\n5+3=8 > 7-YES$

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Answers

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