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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
