Triangle solutions are a crucial topic in both the JEE Main and JEE Advanced exams. The sine rule, cosine rule, tangent rule, and other equations and rules are included in this article. Exam questions frequently include questions based on the application of these formulae. Regularly reviewing these formulae will aid pupils in remembering them and solving problems quickly.
Below is the table of important solutions of triangle formulas for JEE
Important Solutions of Triangle Formulas For JEE
Name of the Formula | Formula |
Sine Rule
| \[\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}\] |
Cosine Rule
| \[cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc}\]
\[cos B = \frac{a^{2} + c^{2} - b^{2}}{2ac}\]
\[cos C= \frac{a^{2} + b^{2} - c^{2}}{2ab}\] |
Projection Rule
| a = b cosC + c CosB
b = a cos C + c Cos A
c = a cos B + b Cos A |
Tangent Rule OR Naiper’s Analogy | \[tan\left ( \frac{B - C}{2} \right ) = \left ( \frac{b - c}{b + c} \right ) cot \frac{A}{2}\]
\[tan\left ( \frac{C - A}{2} \right ) = \left ( \frac{c - a}{c + a} \right ) cot \frac{B}{2}\]
\[tan\left ( \frac{A - B}{2} \right ) = \left ( \frac{a - b}{a + b} \right ) cot \frac{C}{2}\] |
Trignometric Half Angles | \[sin \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{bc}}\]
\[sin \frac{B}{2} = \sqrt{\frac{(s - c)(s - c)}{ac}}\]
\[sin \frac{C}{2} = \sqrt{\frac{(s - a)(s - b)}{ab}}\]
\[cos \frac{A}{2} = \sqrt{\frac{s(s - a)}{bc}}\]
\[cos \frac{B}{2} = \sqrt{\frac{s(s - b)}{ac}}\]
\[cos \frac{C}{2} = \sqrt{\frac{s(s - c)}{ab}}\]
\[tan \frac{A}{2} = \sqrt{\frac{\Delta}{s(s - a)}}\]
\[tan \frac{B}{2} = \sqrt{\frac{\Delta}{s(s - b)}}\]
\[tan \frac{C}{2} = \sqrt{\frac{\Delta}{s(s - c)}}\]
Where, \[\Delta = \sqrt{s(s - a)(s - b)(s - c)}\]
\[s = \frac{a + b + c}{2}\] |
Area of triangles | \[\Delta_{A} = \frac{1}{2}bc sin A\]
\[\Delta_{B} = \frac{1}{2}ac sin B\]
\[\Delta_{C} = \frac{1}{2}ab sin C\] |
M+n Rule | \[(m + n)~cot \theta = m cot\alpha - ncot\beta \] |
Radius of Circumcirle | \[\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} = 2R\]
Where, \[R = \frac{abc}{4 \Delta}\] |
Appolonius Theorem | \[AB^{2} + AC^{2} = 2(AD^{2} + BD^{2})\] |
Radius of the incircle | \[r = \frac{\Delta}{s}\]
\[r = (s - a) tan \frac{A}{2}\]
\[r = (s - b) tan \frac{B}{2}\]
\[r = (s - c) tan \frac{C}{2}\]
\[r = 4R(sin \frac{A}{2})(sin \frac{B}{2})(sin \frac{C}{2})\]
\[cos A + cos B + cos C = 1 + \frac{r}{R}\] |
Radius og the Ex-circle | \[r_{1} = \frac{\Delta}{(s - a)}\]
\[r_{2} = \frac{\Delta}{(s - b)}\]
\[r_{3} = \frac{\Delta}{(s - c)}\]
\[r_{1} = s tan\frac{A}{(2)}\]
\[r_{2} = s tan\frac{B}{(2)}\]
\[r_{3} = s tan\frac{C}{(2)}\]
\[4R = r_{1} + r_{2} + r_{3} + r\] |
Length of angle bisectors | \[x_{A} = \frac{2bc}{b + c} cos\frac{A}{2}\]
\[x_{B} = \frac{2ac}{a + c} cos\frac{B}{2}\]
\[x_{C} = \frac{2ab}{a + b} cos\frac{C}{2}\] |
Length of median | \[l_{A} = \frac{1}{2}\sqrt{2b^{2} + 2c^{2} - a^{2}}\]
\[l_{B} = \frac{1}{2}\sqrt{2a^{2} + 2c^{2} - b^{2}}\]
\[l_{C} = \frac{1}{2}\sqrt{2a^{2} + 2b^{2} - c^{2}}\] |
Length of altitude | \[P_{a} = \frac{2\Delta}{a}\]
\[P_{b} = \frac{2\Delta}{b}\]
\[P_{c} = \frac{2\Delta}{c}\] |
The distances of the special points from vertices and sides of the triangle | Circumcentre \[O_{a} = R cos A\]
\[O_{b} = R cos B\]
\[O_{c} = R cos C\]
Incentre \[I_{A} = r~cosec \frac{A}{2}\]
\[I_{B} = r~cosec \frac{B}{2}\]
\[I_{C} = r~cosec \frac{C}{2}\]
Ex-centre \[I_{1A} = r_{1}~cosec \frac{A}{2}\]
\[I_{1B} = r_{1}~sec \frac{B}{2}\]
\[I_{1C} = r_{1}~sec \frac{C}{2}\]
\[I_{2A} = r_{2}~sec \frac{A}{2}\]
\[I_{2B} = r_{2}~cosec \frac{B}{2}\]
\[I_{2C} = r_{2}~sec \frac{C}{2}\]
\[I_{3A} = r_{3}~sec \frac{A}{2}\]
\[I_{3B} = r_{3}~sec \frac{B}{2}\]
\[I_{3C} = r_{3}~cosec \frac{C}{2}\]
Centroid \[G_{A} = \frac{1}{3} \sqrt{2b^{2} + 2c^{2}} +-a^{2}\]
\[G_{B} = \frac{1}{3} \sqrt{2a^{2} + 2c^{2}} - b^{2}\]
\[G_{C} = \frac{1}{3} \sqrt{2a^{2} + 2b^{2}} - c^{2}\]
Orthocentre Distance of vertices \[H_{A} = 2R cosA\]
\[H_{B} = 2R cosB\]
\[H_{C} = 2R cosC\]
Distance of sides \[H_{A} = 2R cosB cosC\]
\[H_{B} = 2R cosA cosC\]
\[H_{C} = 2R cosA cosB\] |
Solved Examples
Ex.1. In a triangle ABC, the sides are 16 cm, 9 cm, and 8 cm. Find the angle of a triangle for side a.
Solution:
Let a = 16, b = 9, c = 8
First, we want to calculate the s and
\[s = \frac{a + b + c}{2}\]
\[s = \frac{16 + 9 +8}{2}\]
s = 16.5
\[\Delta = \sqrt{s(s - a)(s - b)(s - c)}\]
\[\Delta = \sqrt{16.5(16.5 - 16)(16.5 - 9)(16.5 - 8)}\]
\[\Delta = \sqrt{16.5 \times 0.5 \times 7.5 \times 8.5}\]
\[\Delta = 22.9334\]
\[tan \frac{A}{2} = \frac{\Delta}{s(s - a)}\]
\[tan \frac{120}{2} = \frac{22.9334}{16.5(16.5 - 16)}\]
\[tan 60 = 2.7798\]
Ex.2. Find the length ‘a’ in figure 1, If ∠A = 60°, ∠C = 80° and side c = 5 cm.
Solution:
By using the sine rule we can solve this triangle.
\[\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sinC}\]
\[\frac{a}{sin A} = \frac{c}{sinC}\]
By using cross multiply,
\[a = \frac{sin A}{sin C} \times c\]
\[a = \frac{sin 60}{sin 80} \times 5\]
\[a = 0.8793\times5\]
\[a =4.3969~cm\]
