Law of sines formula for finding sides & angles of triangle

Consider the triangle given below for which we will be writing law of sines formula. In ΔABC, A, B and C are the angles with respective opposite sides a, b and c. In this case, the laws of sine are given by the formula, Thus knowing three parameters and one from each is enough to find the remaining ones as evident from laws of sines.

Law of sines proof for a triangle:

The law of sines proof can be given for all types of triangle. For acute angled triangle, drawing a perpendicular from one of the angle to the opposite side gives a right triangle. Similarly for a obtuse triangle, perpendicular can be drawn and the opposite side can be extended to meet the perpendicular. Finally with the help of right triangle, the law is derived. So lets derive it for a right angled triangle. l

We already know that sine of an angle is equal to the ratio of opposite side to hypotenuse. So we have sin A = a / c, sin B = b / c. => c = a / sinA = b / sin B. Now since this one is a right triangle, sin 90 = 1 or sin C = 1. So c = c sin C or c = c / sin C. Evaluating both the equations, we get the required formula i.e. he law can be extended to derive all this three ratios equal to 2r where r is the circum-radii of the given triangle. This can be called extended law of sines. There is also a condition wherein the law is not able to provide the solution. This occurs when we have measures of 2 sides and a single angle opposite of those. This case is called ambiguous law of sines or rather law of sines ambiguous case. One more variant of the law is spherical law of sines which is valid for spherical triangles. It is as given below. Lets have law of sines examples:

The law of sines examples illustrates the law in a effective manner and makes the topic more informative. Have a look at law of sines example and you'll know it. After solving this, it is advised to create a problem on your own and solve for law of sines practice problems.

(1) Given ΔXYZ, ∠X = 45º, x = 20, y = 10. Find ∠Y:

We have x / sin X = y / sin Y => sin Y = 10 / 20√2 = 1 / 2√2 = 0.35. Hence Y = sin-10.35 = 20.70º. Thus ∠Y is equal to 20.70º.

A look at law of sines word problems:

The law of sines word problems helps to have a mastery over the topic. One such law of sines word problem is given below.

(1) A person is looking at a distant object with angle of elevation 30º. The angle between line of sight and perpendicular from object is 60º. Find the linear distance of object if its height is 100 m:

We know that the height is opposite to angle of elevation and object angle being opposite to distance, we have D / sin 60 = 100 / sin 30 => D = 100 sin 60 / sin 30 = 100 x (√3/2) / (½) = 100 x √3. Thus the object is at a distance of 173.2 m on x-axis.

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