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Veronika [31]
3 years ago
14

WHERE DOES THE PYTHAGOREAN IDENTITY SIN2 Θ + COS2 Θ = 1 COME FROM? HOW WOULD YOU USE IT TO FIND THE SINE COSINE AND TANGENT VALU

ES OF THE ANGLE?
Mathematics
2 answers:
Allushta [10]3 years ago
7 0

Imagine a trigonometric circle with radius equal to 1.

We can say that the opposite side of the angle generated by the origin is equal to sen Ф and the adjacent equal to cos Ф


With this we can say that:


h² = c² + c²


If r = 1, then h = 1


1² = cos² + sen²

1 = cos² Ф + sen² Ф


You can use this equation a lot of times because it's the fundamental trigonometric relation
, so, when you something like:


sen Ф = cos Ф + 1


you can take from sen² Ф + cos² Ф = 1 that sen = \sqrt{1 - cos^2} and then resolve.

lesya [120]3 years ago
4 0

Suppose we have a right- angled triangle with theta has one of the angles ( not 90 degrees) and hypotenuse c , opposite side a and adjacent side b.

then sin α = a/c giving a = c sin α .............(1)

and cos α = b/c giving b = c cos α...............(2)

by Pythagoras theorem:-

a^2 + b^2 = c^2 and from equations (1) and (2):-

a^2 + b^2 = c^2 sin^2 α + c^2 cos^2 α

a^2 + b^2 = c^2 ( sin^2 α + cos^2 α)

Comparing this equation with the Pythagoras equation sin^2 α + cos^2 α must equal 1.

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Please solve with explanation (high points)
Hunter-Best [27]

Step-by-step explanation:

so, we have a large triangle made of the 2 cables as legs and the ground distance AB as baseline.

the tower is the height to the baseline of that large triangle.

let's call the top of the tower T.

and remember, the sum of all angles in a triangle is always 180°.

we know the angle A = 62°, and angle B = 72°.

assuming that AB is a truly horizontal line that means that the 2 legs (cables) have different lengths, the triangle is not isoceles, and the tower is not in the middle of the baseline.

so, the height (tower) splits the baseline into 2 parts. let's call them p and q.

p + q = 12 m

p = 12 - q

let's simply define that p is the part of the baseline on the A side, and q is the part of the baseline on the B side.

we have now 2 small right-angled triangles the large height (tower) splits the large triangle into.

one has the sides

AT, height (tower), p

angle A = 62°

angle T = 180 - 90 - 62 = 28°

the other has the sides

BT, height (tower), q

angle B = 72°

angle T = 180 - 90 - 72 = 18°

now remember the law of sine :

a/sin(A) = b/sin(B) = c/sin(C)

with the sides and the associated angles being opposite.

p/sin(28) = height/sin(62)

q/sin(18) = height/sin(72)

we know from above that

p = 12 - q

so,

(12 - q)/sin(28) = height/sin(62)

height = (12 - q)×sin(62)/sin(28)

q/sin(18) = height/sin(72)

height = q×sin(72)/sin(18)

and therefore, as height = height we get

(12 - q)×sin(62)/sin(28) = q×sin(72)/sin(18)

(12 - q)×sin(62)×sin(18) = q×sin(72)×sin(28)

12×sin(62)×sin(18) - q×sin(62)×sin(18) =

= q×sin(72)×sin(28)

12×sin(62)×sin(18) = q×sin(72)×sin(28) + q×sin(62)×sin(18) =

= q×(sin(72)×sin(28) + sin(62)×sin(18))

q = 12×sin(62)×sin(18) / (sin(72)×sin(28) + sin(62)×sin(18))

q = 4.551603755... m

p = 12 - q = 7.448396245... m

height = q×sin(72)/sin(18) = 14.00839594... m ≈ 14 m

the cell tower is about 14 m tall.

7 0
1 year ago
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