The document Notes | EduRev is a part of the JEE Course JEE Main Mock Test Series 2020 & Previous Year Papers.

All you need of JEE at this link: JEE

**Sine rule: Sides of a triangle are proportional to the sine of the angles opposite to them. So, in Î”ABC,**

sin A/a = sin B/b = sin C/c = 2Î”/abc.

This may also be written as (a/sin A) = (b/sin B) = (c/sin C)**Cosine rule: In any Î”ABC,**

cos A = (b^{2}+ c^{2}â€“ a^{2}) /2bc

cos B = (a^{2}+ c^{2}â€“ b^{2})/2ac

cos C = (a^{2}+ b^{2}- c^{2})/2ab**Trigonometric ratios of half-angles:**

sin A/2 = âˆš[(s-b)(s-c)/bc]

sin B/2 = âˆš[(s-c) (s-a)/ac]

sin C/2 = âˆš[(s-a) (s-b)/ab]

cos A/2 = âˆšs(s - a)/bc

cos B/2 = âˆšs(s - b)/ac

cos C/2 = âˆšs(s - c)/ab

tan A/2 = âˆš[(s - b) (s - c)/s(s - a)]

tan B/2 = âˆš[(s - c) (s - a)/s(s - b)]

tan C/2 = âˆš[(s - a) (s - b)/s(s - c)]**Projection rule: In any Î”ABC,**

a = b cos C + c cos B

b = c cos A + a cos C

c = a cos B + b cos A**Area of a triangle**

If Î” denotes the area of the triangle ABC, then it can be calculated in any of the following forms:

Î” = 1/2 bc sin A = 1/2 ca sin B = 1/2 ab sin C

Î” = âˆšs(s - a)(s â€“ b)(s - c)

Î” = 1/2. (a^{2}sin B sin C)/ sin(B + C)

= 1/2. (b^{2}sin C sin A)/ sin (C + A)

= 1/2. (c^{2}sin A sin B)/ sin (A + B)**Semi-perimeter of the triangle**

If S denotes the perimeter of the triangle ABC, then s = (a + b + c)/2**Napierâ€™s analogy**

In any Î”ABC,

tan [(B â€“ C)/2] = (b â€“ c)/(b + c) cot A/2

tan [(C â€“ A) /2] = (c â€“ a)/(c + a) cot B/2

tan [(A â€“ B) /2] = (a â€“ b)/(a + b) cot C/2**m-n theorem**

Consider a triangle ABC where D is a point on side BC such that it divides the side BC in the ratio m: n, then as shown in the figure, the following results hold good:

(m + n) cot Î¸ = m cot Î± â€“ n cot ÃŸ.

(m + n) cot Î¸ = n cot B â€“ m cot C.**Apollonius theorem**

In a triangle ABC, if AD is the median through A, then

AB^{2 }+ AC^{2}= 2(AD^{2}+ BD^{2}).- If the three sides say a, b and c of a triangle are given, then angle A is obtained with the help of the formula

tan A/2 = âˆš(s - b) (s - c) / s(s - a) or cos A = b^{2}+ c^{2}- a^{2}/ 2bc.

Angles B and C can also be obtained in the same way. - If two sides b and c and the included angle A are given, then

tan (B â€“ C) /2 = (b â€“ c)/ (b + c) cot A/2

This gives the value of (B- C)/2.

Hence, using (B + C)/2 = 90^{o}- A/2 along with the last equation both B and C can be evaluated. Now, the sides can be evaluated using the formula

a = b sin A/sin B or a^{2}= b^{2}+ c^{2}â€“ 2bc cosA. - If two sides b and c and the angle B (opposite to side b) are given, then using the following results, we can easily obtain the remaining elements

sin C = c/b sinB, A = 180^{o }â€“ (B + C) and b = b sin A/sinB - Some of the important points which must be remembered include:

1. b < c sin B, there is no triangle possible (Fig. 1)

2. If b = c sin B and B is an acute angle, then only one triangle is possible (Fig. 2)

3. If c sin B < b < c and B is an acute angle, then there are two values of angle C (Fig. 3).

4. If c < b and B is an acute angle, then there is only one triangle (Fig. 4).

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

3 videos|174 docs|149 tests

### Solved Examples - Solution of Triangles

- Doc | 2 pages
### Revision Notes - Circles Connected with Triangles

- Doc | 2 pages
### Solved Examples - Circles Connected with Triangles

- Doc | 2 pages
### Revision Notes - Trigonometric Functions

- Doc | 3 pages
### Solved Examples - Trigonometric Functions

- Doc | 1 pages