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svetoff [14.1K]
2 years ago
10

Question. 1 :

Mathematics
1 answer:
vovikov84 [41]2 years ago
4 0
<h3>Answer to Question 1:</h3>

AB= 24cm

BC = 7cm

<B = 90°

AC = ?

<h3>Using Pythagoras theorem :-</h3>

AC^2 = AB^2 + BC ^ 2

AC^2 = 24^2 + 7^2

AC^2 = 576 + 49

AC^2 = √625

AC = 25

<h3>Answer to Question 2 :-</h3>

sin A = 3/4

CosA = ?

TanA = ?

<h3>SinA = Opp. side/Hypotenuse</h3><h3> = 3/4</h3>

(Construct a triangle right angled at B with one side BC of 3cm and hypotenuse AC of 4cm.)

<h3>Using Pythagoras theorem :-</h3>

AC^2 = AB^2 + BC ^ 2

4² = AB² + 3²

16 = AB + 9

AB = √7cm

<h3>CosA = Adjacent side/Hypotenuse</h3>

= AB/AC

= √7/4

<h3>TanA= Opp. side/Adjacent side</h3>

=BC/AB

= 3/√7

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The course for a boat race starts at point A and proceeds in the direction of S52'E for 1 hour at 8 knots to point B and then in
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Answer:

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Step-by-step explanation:

I assume you mean ° (degrees), not ' (minutes).  There are 60 minutes in 1 degree.

S52°E means "south, 52° east", or 52° east of south.

S50°W means "south, 50° west", or 50° west of south.

1 knots = 1 nautical mile / hour, so the boat first travels 8 nautical miles from A to B, then 4.5 nautical miles from B to C, then finally back to A.

If we say A is at the origin, then the coordinates of B are:

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And the coordinates of C are:

(8 sin 52° − 4.5 sin 50°, -8 cos 52° − 4.5 cos 50°)

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So the distance from A to C is:

x = √(2.857² + (-7.818)²)

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And the total distance of the race is:

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The compass bearing from A to C is:

θ = atan(2.857 / 7.818)

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