Q: What is the cos of angle 'a' with the measurements 5 13 12?

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It is down to trigonometry. You have two parts of a triangle, so you are looking for the length of the last line. As the man turned right, there will be a right angle in the triangle. As we have two sides, the 12m and 5m, using Pythagoras theorem we can work out the length of the last side. The square of the hypotenuse is equal to the square of the other two sides. 12 x 12 = 144. 5 x 5 = 25. 144 + 25 = 169. 13 x 13 = 169. So the length of the other side is 13.

13 feet

It is 3/13 - 2/13*i

5/13

You only have to be 13 but at 13 you need a parent.All piercings hurt to some degree. For me, the tragus was just a pinch and then afterwards, there was a burning sensation for about an hour. Everyone is different though.

Related questions

It is: cos^-1(12/13) = 22.61986495 degrees

It is: cos = adj/hyp and the acute angles for the given right angle triangle are 67.38 degrees and 22.62 degrees

It is about 67 degrees

The complement of an acute angle A is the angle 90° - A. The complement of 13° is 77°.

The dimensions given fits that of a right angle triangle and sin^-1(12/13) = 67.38 degrees

It might be pythagoras therom but it can only be Pythagoras when the traingle has a right angle. If it does then try to work it out using phythagoras. If the angle between the given sides is B, then: b2 = a2 + c2 - 2ac cos B ⇒ b2 = (7 cm)2 + (13 cm)2 - 2 x 7 cm x 13 cm x cos B ⇒ b = √(218 - 182 cos B) cm If it is a right angle triangle, with B the right angle, cos B = cos 90o = 0 and this becomes Pythagoras making the side: b = √218 cm ≈ 14.76 cm If there is a right angle, not between the 7 cm and 13 cm, then the 13cm side is the hypotenuse (as the hypotenuse must be the longest side) and the other side is: b = √(132 - 72) cm = √120 cm ≈ 10.95 cm

If you mean 5 by 12 by 13 then they will form a right angle triangle

sin = -12/13 cos = 5/12 tan = -5/12 cosec = -13/12 sec = 12/5 cotan = -12/5

You have not indicated which side the angle is opposite of. Can us law of cosines then by calling sides c and a. b^2 = a^2 + c^2 - 2(a)(c) cos(B) I would arbitrarily have to assign values you have not given me.

Writing x instead of theta, cos2x = 1 - (12/13)2 = 1 - 144/169 = 25/169 = (5/13)2 So cos(x) = Â± 5/13 so that x = cos-1(5/13) or cos-1(-5/13) And then, depending on the range of x, you have solutions for x. A calculator will only give you the principal solutions, though.

If the right angle is at A then SA = 5 mm.

5/13 = 0.3846 (to 4 dp)