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Find angle B in triangle ABC if a is 10 b 16 and c 12?
We can use the law of cosines here. ( remember, side b is opposite angle B) DEGREE MODE! b^2 = a^2 + c^2 - 2ac cos(B) 16^2 = 10^2 + 12^2 - 2(10)(12) cos(B) 256 = 244 - 240(cos B ) 12 = -240(cos B ) -0.05 = cosB arcos(-0.05) = B B = 93 degrees
What is the law of sines and the law of cosines and how do they help solve a triangles' angles and sides?
In a triangle ABC, with side a opposite angle A, side b opposite angle B and side c opposite angle C, the sine rule is: sin(A)/a = sin(B)/b = sin(C)/c The cosine rule is: a2 = b2 + c2 - 2bc*cos(A) and, by symmetry, b2 = c2 + a2 - 2ca*cos(B) c2 = a2 + b2 - 2ab*cos(C)
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
How do you find unknown angles if you know 3 sides?
For a triangle with sides a, b anc c, where A is the angle opposite side a, B is the angle opposite side b, etc.:cos A = ( b2 + c2 - a2 ) / ( 2 bc )cos B = ( a2 + c2 - b2 ) / ( 2 ac )cos C = ( a2 + b2 - c2 ) / ( 2 ab )These are just rearrangements of the ordinary cosine rule:a2 = b2 + c2 - 2 bc cos A
If the cosine of angle B is 0.2536 in a right triangle how can I find the measure of ( acute ) angle B with just that information?
Angle of B is cos^-1*(0.2536) = 75.309 degrees to three decimal places
What is this expression as the cosine of an angle cos30cos55 plus sin30sin55?
cos(30)cos(55)+sin(30)sin(55)=cos(30-55) = cos(-25)=cos(25) Note: cos(a)=cos(-a) for any angle 'a'. cos(a)cos(b)+sin(a)sin(b)=cos(a-b) for any 'a' and 'b'.
I have magnitude and angle how can find a b and a Dot b a 3 angle 45 b 5 angle 90 Find A B A dot B?
a = 3/sqrt(2)*i + 3/sqrt(2)*jb = 5ja.b = |a|*|b|*cos(q)= 3*5*cos(45) = 15/sqrt(2)
What is cos of a negative angle?
The cosine function is an even function which means that cos(-x) = cos(x). So, if cos of an angle is positive, then the cos of the negative of that angle is positive and if cos of an angle is negative, then the cos of the negative of that angle is negaitive.
Find angle B in triangle ABC if a is 10 b 16 and c 12?
We can use the law of cosines here. ( remember, side b is opposite angle B) DEGREE MODE! b^2 = a^2 + c^2 - 2ac cos(B) 16^2 = 10^2 + 12^2 - 2(10)(12) cos(B) 256 = 244 - 240(cos B ) 12 = -240(cos B ) -0.05 = cosB arcos(-0.05) = B B = 93 degrees
Express cos4x sin3x in a series of sines of multiples of x?
The best way to answer this question is with the angle addition formulas. Sin(a + b) = sin(a)cos(b) + cos(a)sin(b) and cos(a + b) = cos(a)cos(b) - sin(a)sin(b). If you compute this repeatedly until you get sin(3x)cos(4x) = 3sin(x) - 28sin^3(x) + 56sin^5(x) - 32sin^7(x).
When cross and dot product equal what is angle between A and B?
A · B = |A| |B| cos(Θ)A x B = |A| |B| sin(Θ)If [ A · B = A x B ] then cos(Θ) = sin(Θ).Θ = 45°
How do you find hypotenuse c angle a equals 30 deg side b equals 5 mm?
To find the hypotenuse with angle a and side b, we use the identity below:cos(a) = b/cWe have a and b, and to find c, we multiply both sides by c and divide both sides by cos(a):c = b/cos(a)c = 5/cos(30)c = 32.41460617mm
What is the meaning of sin cos tan?
There is more than one equivalent definition. One way to think of this is this: Imagine a right triangle, with an angle "x". The sides of the triangles are as follow: side "a" is opposite to the angle "x", side "b" is adjacent to the angle, and side "c" is the hypothenuse (the longest side, opposite the right angle). In this case: * sin(x) = a/c * cos(x) = b/c * tan(x) = a/b = sin(x) / cos(x) * cot(x) = b/a = cos(x) / sin(x) * csc(x) = c/a = 1 / sin(x) * sec(x) = c/b = 1 / cos(x) These ratios of sides will depend on the angle "x", but for any angle "x", the ratio will always be the same. For example, for an angle of 30°, the ratio a/c (sine of x) will always be 1/2.
How do you solve double angle equations for trigonometry?
There are two ways to solve for the double angle formulas in trigonometry. The first is to use the angle addition formulas for sine and cosine. * sin(a + b) = sin(a)cos(b) + cos(a)sin(b) * cos(a + b) = cos(a)cos(b) - sin(a)sin(b) if a = b, then * sin(2a) = sin(a)cos(a) + cos(a)sin(a) = 2sin(a)cos(a) * cos(2a) = cos2(a) - sin2(b) The cooler way to solve for the double angle formulas is to use Euler's identity. eix = cos(x) + i*sin(x). Yes, that is "i" as in imaginary number. we we put 2x in for x, we get * e2ix = cos(2x) + i*sin(2x) This is the same as * (eix)2 = cos(2x) + i*sin(2x) We can substitute our original equation back in for eix. * (cos(x) + i*sin(x))2 = cos(2x) + i*sin(2x) We can distribute the squared term. * cos2(x) + i*sin(x)cos(x) + i*sin(x)cos(x) + (i*sin(x))2 = cos(2x) + i*sin(2x) And simplify. Because i is SQRT(-1), the i squared term becomes negative. * cos2(x) + 2i*sin(x)cos(x) - sin2(x) = cos(2x) + i*sin(2x) * cos2(x) - sin2(x) + 2i*sin(x)cos(x) = cos(2x) + i*sin(2x) Now you can plainly see both formulas in the equation arranged quite nicely. I don't yet know how to get rid of the i, but I'm working on it.
How do you find the base of a triangle if is not given?
You use trigonometry. If the triangle is a right triangle, then you can use the Pythagorean theorem (a2 + b2 = c2 where c is the hypotenuse). This requires you to know two of the sides of the triangle. You can also use the relationship: sin A = a/c cos A = b/c tan A = a/b where "A" is a non-right angle of a right triangle, "a" is the length of the side opposite of the angle "A", "b" is the length of the side adjacent to the angle "A" and "c" is the length of the hypotenuse. If the triangle is NOT a right triangle, you can use the law of sines or the law of cosines. The law of sines: a /sin A = b / sin B = c / sin C where "a" is the side opposite of angle "A", "b" is the side opposite of angle "B" and "c" is the side opposite of angle "C". The law of cosines: a2 = b2 + c2 - b*c*cos A b2 = a2 + c2 - a*c*cos B c2 = a2 + b2 - a*b*cos C where "c" is the hypotenuse, "a" and "b" are the other sides of the triangle and "C" is the angle opposite of "c", "B" is the angle opposite of "b" and "A" is the angle opposite of "a".
What is the law of sines and the law of cosines and how do they help solve a triangles' angles and sides?
In a triangle ABC, with side a opposite angle A, side b opposite angle B and side c opposite angle C, the sine rule is: sin(A)/a = sin(B)/b = sin(C)/c The cosine rule is: a2 = b2 + c2 - 2bc*cos(A) and, by symmetry, b2 = c2 + a2 - 2ca*cos(B) c2 = a2 + b2 - 2ab*cos(C)
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
