0
Use Cosine Rule
a^(2) = b^(2) + c^(2) - 2bcCosA
Algebrically rearrange
CosA = [a^(2) - b^(2) - c^(2)] / -2bc
Substitute
CosA = [13^(2) - 12^(2) - 5^(2)# / -2(12)(5)
CosA = [ 169 - 144 - 25] / -120
Cos)A) = [0] / -120
CosA = 0
A = 90 degrees (the right angle opposite the hypotenuse)/
However,
If 'A' is the angle between '12' & '13' then 'a' is the side '5'
Hence (Notice the rearrangement of the numerical values).
CosA = [5^(2) - 12^(2) - 13^(2) ] / -2(12)(13)
CosA = [ 25 - 144 -169] / -312
CosA = [ -288[/-312
CosA = 288/312
A = Cos^(-1) [288/312]
A = 22.61986495.... degrees.
What else can I help you with?
What is the cos of angle B 5 13 12?
To find the cosine of angle B given the sides of a triangle, you typically use the cosine rule or the relationship between the sides. However, the values "5," "13," and "12" seem to refer to the lengths of the sides of a triangle. If these correspond to a triangle with sides a = 5, b = 12, and c = 13, you can use the cosine rule: ( \cos(B) = \frac{a^2 + c^2 - b^2}{2ac} ). Plugging in the values, ( \cos(B) = \frac{5^2 + 13^2 - 12^2}{2 \cdot 5 \cdot 13} = \frac{25 + 169 - 144}{130} = \frac{50}{130} ), which simplifies to ( \cos(B) = \frac{5}{13} ).
What is the sin of angle B if the angles are 5 12 and 13?
This is a classic Pythagorean triangle. Although you have given the side lengths, you have NOT given a letter to correspond , with the given side. However, Let 12 be the adjacentr side (base) Let '5' be the opposite side ( perpendicular ) Let '13' by the hypotenuse. Sin(Angle) = opposite / hypotenuse = 5/13 Angle = Sin^(-1) 5/13 = 22.619... degrees. NB This is the angle between the hypotenuse and the base(adjacent) Now 'swopping' things around , we take the angle between the hypotenuse and the perpendicular (opposite) . This now becomes perpendicular(adjacent) and the base becomes the opposite. Hence Sin(angle) = 12/13 Angle = Sin^(-1) 12/13 = 67.380.... degrees. The angle at the 'top' of the triangle. Verification. ' 90 + 67.380... + 22.619... = 180 ( allow for calculator decimals).
What is the tan of angle B in a 12 by 13 by 5 triangle?
In a triangle with sides measuring 12, 13, and 5, we can identify the angle opposite the side measuring 5 as angle B. To find the tangent of angle B, we use the formula ( \tan(B) = \frac{\text{opposite}}{\text{adjacent}} ). Here, the side opposite angle B is 5, and the adjacent side (which can be either of the other two sides depending on which angle we consider) is 12. Therefore, ( \tan(B) = \frac{5}{12} ).
When a man walks 5m towards east and turns right and moves 12m the magnitude of his displacement is 13m how?
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.
The foot of a ladder is placed 5 feet from a building the top of the ladder rests 12 feet up on the building how long is the ladder?
13 feet
What is the cos of angle a if the opposite is 5 the adjacent is 12 and the hypotenuse is 13?
It is: cos^-1(12/13) = 22.61986495 degrees
What is cos of angle a with a hypotenuse of 13 and adjacent of 5 and a opposite of 12?
Cos(angle) = adjacent / hypotenuse. Cos(a) = a/h Substitute Cos(X) = 5/13 = 0.384615... A = Cos^*-1( 0.384615 .... A = 67.38013505... degrees.
What is the cos of angle a 5 13 12?
It is: cos = adj/hyp and the acute angles for the given right angle triangle are 67.38 degrees and 22.62 degrees
What is the cos of angle B 5 13 12?
To find the cosine of angle B given the sides of a triangle, you typically use the cosine rule or the relationship between the sides. However, the values "5," "13," and "12" seem to refer to the lengths of the sides of a triangle. If these correspond to a triangle with sides a = 5, b = 12, and c = 13, you can use the cosine rule: ( \cos(B) = \frac{a^2 + c^2 - b^2}{2ac} ). Plugging in the values, ( \cos(B) = \frac{5^2 + 13^2 - 12^2}{2 \cdot 5 \cdot 13} = \frac{25 + 169 - 144}{130} = \frac{50}{130} ), which simplifies to ( \cos(B) = \frac{5}{13} ).
If sin A 513 for angle A in Quadrant I find cos 2A?
To find ( \cos 2A ) using the given ( \sin A = \frac{5}{13} ), we first use the Pythagorean identity to find ( \cos A ). Since ( \sin^2 A + \cos^2 A = 1 ), we have ( \cos^2 A = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169} ). Thus, ( \cos A = \frac{12}{13} ). Using the double angle formula ( \cos 2A = 2\cos^2 A - 1 ), we get ( \cos 2A = 2\left(\frac{12}{13}\right)^2 - 1 = 2 \cdot \frac{144}{169} - 1 = \frac{288}{169} - \frac{169}{169} = \frac{119}{169} ).
What is the complement of an angle whose measure is 13 degrees?
The complement of an acute angle A is the angle 90° - A. The complement of 13° is 77°.
What is the sin of angle A 5 13 12?
The dimensions given fits that of a right angle triangle and sin^-1(12/13) = 67.38 degrees
What is the length of missing side b in the triangle 7cm and 13 cm?
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
What is the exact ratio of all 6 trig functions when p is a point on a circle at 5 -12?
sin = -12/13 cos = 5/12 tan = -5/12 cosec = -13/12 sec = 12/5 cotan = -12/5
What is the measurement of angle B in the triangle when the hypotenuse is 13 the opposite side is 12 and the adjacent side is 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.
How do you calculate cos squared theta equals one minus twelve thirteenths squared?
Writing x instead of theta, cos2x = 1 - (12/13)2 = 1 - 144/169 = 25/169 = (5/13)2 So cos(x) = ± 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.
What is the sin of angle B if the angles are 5 12 and 13?
This is a classic Pythagorean triangle. Although you have given the side lengths, you have NOT given a letter to correspond , with the given side. However, Let 12 be the adjacentr side (base) Let '5' be the opposite side ( perpendicular ) Let '13' by the hypotenuse. Sin(Angle) = opposite / hypotenuse = 5/13 Angle = Sin^(-1) 5/13 = 22.619... degrees. NB This is the angle between the hypotenuse and the base(adjacent) Now 'swopping' things around , we take the angle between the hypotenuse and the perpendicular (opposite) . This now becomes perpendicular(adjacent) and the base becomes the opposite. Hence Sin(angle) = 12/13 Angle = Sin^(-1) 12/13 = 67.380.... degrees. The angle at the 'top' of the triangle. Verification. ' 90 + 67.380... + 22.619... = 180 ( allow for calculator decimals).
