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The subject of trigonometry has a bearing on triangles. In fact the name trigonometry is derived combining the terms triangles geometry. Each of the basic trigonometric ratios (Sine, Cosine and Tangents) of the interior angles of any triangle have some established relations with the sides of the triangle. Some important relations out of those are framed as trigonometry laws of a triangle or in short as trigonometry laws. Let us study those important laws one by one.

In any scalene triangle ABC , as shown above, the law of sine in the form of formula is,

$[\frac{(a)}{(\sin A)}]$ = $[\frac{(b)}{(\sin B)}]$ = $[\frac{(c)}{\sin C}]$The above formula is known as law of sines and let us show how this law is established.

In the same triangle draw the altitude h which divides a into two parts as d and e.

Now, h = c*sin B from left side of A and h = b*sin C

Therefore, b*sin C = c*sin B or, $\frac{(b)}{(sin B)}$ = $\frac{(c)}{(sin C)}$

In the same way, by drawing a perpendicular from B on AC, it can be shown that

$\frac{(a)}{(sin A)}$ = $\frac{(c)}{(sin C)}$

Therefore, $\frac{(a)}{(sin A)}$ = $\frac{(b)}{(sin B)}$ = $\frac{(c)}{(sin C)}$

Thus the law of sine is proved.

In the same triangle draw the altitude h which divides a into two parts as d and e.

Let us prove the law of cosine with the same diagram

a = d + e = c cos B + b cos C. Multiplying by a both sides,

a

b

c

b

or, a

b

Thus the law of cosine is proved.

Suppose we only know the measures of two sides and one angle which is not an included angle. In this case, the entire triangle can be solved by law of sine alone. But in some cases, there is a possibility of getting two solutions or ambiguous solutions. Both the solutions may be correct or one of them may be extraneous depending upon the application. Let us explain cases where two possible solutions can occur.

Consider the case when the measures of sides b, c and measure of angle B is known. There will be two solutions for angle A correspondingly two solutions for the measures of side a, if,

1) angle B is acute

2) b < c

3) b > c*sin B

Similarly law of cosine is ineffective if we only know the measures of two sides and one angle which is not an included angle. In this case, the entire triangle can be solved by law of sine alone.

The law of cosine is also helpful to find the type of triangle when you know only the measures of all the sides. The method is, label the greatest side as a, thereby, the greatest angle is A. Now rewriting the formula a

$\cos A$ = $\frac{[(b^{2}+c^{2})-(a^{2})]}{[(2bc )]}$If cos A is positive, the angle A is acute, and hence the triangle is an acute triangle as A is the greatest angle. If cos A is negative, the angle A is obtuse, and hence the triangle is an obtuse triangle as A is the greatest angle. If cos A is 0, the angle A is right angle, and hence the triangle is a right triangle.

$\frac{[(a-b)]}{[(a+b)]}$ = $[\frac{\tan \frac{1}{2}(A-B)}{\tan \frac{1}{2}(A+B)}]$The law of tangents is proved as follows.

As per law of sines,

$\frac{a}{\sin A}$ = $\frac{b}{\sin B}$. Let this ratio be k.

Then, a = k*sin (A) and b = k*sin (B)

Therefore,

$\left [ \frac{(a-b)}{(a+b)} \right ]$ = $\frac{[k*\sin (A)-k*\sin (B)]}{[k*\sin (A)+k*\sin (B)]}$ = $\frac{[\sin (A)-\sin (B)]}{[\sin (A)+\sin (B)]}$

As per the formula for sum and difference of two sine ratios,

sin (A) - sin (B) = 2 sin $\frac{1}{2}$ (A – B)*cos $\frac{1}{2}$ (A + B) and sin (A) + sin (B) = 2 sin $\frac{1}{2}$ (A + B)*cos $\frac{1}{2}$ (A - B)

Therefore,

$\left [ \frac{(a-b)}{(a+b)} \right ]$ = $\frac{[2\sin \frac{1}{2}(A-B)*\cos \frac{1}{2}(A+B)]}{2\sin \frac{1}{2}(A+B)*\cos \frac{1}{2}(A-B)}$

=$\frac{[\tan \frac{1}{2}(A-B)]}{[\tan \frac{1}{2}(A+B)]}$

Thus the law of tangent is proved.

This law is more convenient than the law of cosine to solve a triangle when only two sides and the included angle is known. It is explained below. The above formula can be rewritten as,

$\tan$ $\frac{1}{2}$ (A – B) = $[\frac{(a – b)}{(a + b)}]$ * $\tan$ $\frac{1}{2}$ (A + B) = $[\frac{(a – b)}{(a + b)}]$ * $\cot$ $\frac{1}{2}$ (C)Since we know all the items on the right, (A – B) can be figured out.

We also know (A + B), since (A + B) = 180

By solving we can find the angles A and B. The remaining side c can now be found by law of sine.