Important Solutions of Triangle Formulas For JEE Main and Advanced

Solutions of Triangle is an important topic in the JEE Main and JEE Advanced. This topic comprises various formulae and rules like the sine rule, cosine rule, tangent rule etc. Questions based on the application of these formulas are often asked in exams. Revising these formulas on a regular basis will help students to remember them and easily solve the questions.

Solutions of Triangle Formulas

1. Sine Rule:

a/sin A = b/sin B = c/sin C

2. Cosine Formula:

In any triangle ABC,

(i) cos A = (b2+c2-a2)/2bc

(ii) cos B = (c2+a2-b2)/2ca

(iii) cos C = (a2+b2-c2)/2ab

3. Projection Formula:

In any triangle ABC,

(i) a = b cos C + c cos B

(ii) b = c cos A + a cos C

(iii) c = a cos B + b cos A

4. Napier’s Analogy- Tangent rule:

(i) tan(BC2)=(bcb+c)cotA2\tan \left ( \frac{B-C}{2} \right ) = \left ( \frac{b-c}{b+c} \right )\cot \frac{A}{2}

(ii) tan(CA2)=(cac+a)cotB2\tan \left ( \frac{C-A}{2} \right ) = \left ( \frac{c-a}{c+a} \right )\cot \frac{B}{2}

(iii) tan(AB2)=(aba+b)cotC2\tan \left ( \frac{A-B}{2} \right ) = \left ( \frac{a-b}{a+b} \right )\cot \frac{C}{2}

5. Trigonometric functions of half angles:

(i) sinA2=(sb)(sc))bc\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c))}{bc}}

sinB2=(sc)(sa))ca\sin \frac{B}{2} = \sqrt{\frac{(s-c)(s-a))}{ca}}

sinC2=(sa)(sb))ab\sin \frac{C}{2} = \sqrt{\frac{(s-a)(s-b))}{ab}}

(ii) cosA2=s(sa)bc\cos \frac{A}{2} = \sqrt{\frac{s(s-a)}{bc}}

cosB2=s(sb)ac\cos \frac{B}{2} = \sqrt{\frac{s(s-b)}{ac}}

cosC2=s(sc)ab\cos \frac{C}{2} = \sqrt{\frac{s(s-c)}{ab}}

(iii) tanA2=(sb)(sc)s(sa)=Δs(sa)\tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \frac{\Delta }{s(s-a)}

where s = (a+b+c)/2 , the semi perimeter of the triangle.

(iv) sin A = 2bcs(sa)(sb)(sc)=2Δbc\frac{2}{bc}\sqrt{s(s-a)(s-b)(s-c)} = \frac{2\Delta }{bc}

6. Area of Triangle(Δ) = (½) ab sin C = (½) bc sin A = (½) ca sin B = s(sa)(sb)(sc)\sqrt{s(s-a)(s-b)(s-c)}

7. M-n rule:

If D be the point on the side BC of a triangle ABC which divides the side BC in the ratio m: n,

If BD:DC = m:n, ∠ BAD = α, ∠ CAD = β, ∠ CDA = θ, then 

(m+n) cot θ = m cot α- n cot β = n cot B – m cot C

8. Radius of circumcircle:

R = a/2sinA = b/2sinB = c/2sinC = abc/4Δ 

9. Apollonius theorem: In a triangle ABC, if AD is the median through A, then

AB2 + AC2 = 2(AD2 + BD2).

10. Radius of the incircle:

(i) r = Δ/s

(ii) r = (s-a) tan (A/2) = (s-b) tan(B/2) = (s-c) tan(C/2)

(iii) r = asinB2sinC2cosA2\frac{a\sin \frac{B}{2}\sin \frac{C}{2}}{\cos \frac{A}{2}} and so on.

(iv) r = 4R sin (A/2) sin (B/2) sin (C/2)

11. Radius of the Ex-circles:

(i) r1 = Δ/(s-a)

r2 = Δ/(s-b)

r3 = Δ/(s-c)

(ii) r1 = s tan (A/2)

r2 = s tan (B/2)

r3 = s tan (C/2)

(iii) r1 = acosB2cosC2cosA2\frac{a\cos \frac{B}{2}\cos \frac{C}{2}}{\cos \frac{A}{2}} and so on.

(iv) r1 = 4R sin (A/2) cos (B/2) cos (C/2)

12. Length of angle bisectors, medians and altitudes:

(i) Length of an angle bisector from angle A = βa = 2bccosA2b+c\frac{2bc\: \cos \frac{A}{2}}{b+c}

(ii) Length of median from angle A = ma = 122b2+2c2a2\frac{1}{2}\sqrt{2b^{2}+2c^{2}-a^{2}}

(iii) Length of altitude from angle A = Aa = 2Δ/a

13. The distances of the special points from vertices and sides of triangle:

(i) circumcentre (O) : OA = R and Oa= R cos A

(ii) Incentre (I): IA = r cosec(A/2) and Ia = r

(iii) Excentre (I1): I1A = r1 cosec(A/2)

(iv) Orthocentre: HA = 2R cos A and Ha = 2R cos B cos C

(v) Centroid (G): GA = 132b2+2c2a2\frac{1}{3}\sqrt{2b^{2}+2c^{2}-a^{2}} and Ga = 2Δ/3a

14. Orthocentre and Pedal Triangle:

The triangle formed by joining the feet of the altitudes is called the Pedal Triangle.

(i) Its angles are π – 2A, π – 2B and π – 2C.

(ii) The sides are a cos A = R sin 2A

a cos B = R sin 2B

a cos C = R sin 2C

(iii)Circum radii of the triangle PBC, PCA, PAB and ABC are equal.

15. Excentral Triangle:

The triangle formed by joining the three excentres I1, I2 and I3 of ΔABC is called the excentral triangle.

(i)ΔABC is the pedal triangle of the ΔI1I2I3.

(ii) Angles are π2A2\frac{\pi }{2}-\frac{A}{2} , π2B2\frac{\pi }{2}-\frac{B}{2} and π2C2\frac{\pi }{2}-\frac{C}{2}.

(iii) Sides are 4R cos (A/2), 4R cos (B/2) and 4R cos (C/2)

(iv) I I1 = 4R sin(A/2), I I2 = 4R sin(B/2), I I3 = 4R sin(C/2) 

(v)Incentre I of ΔABC is the orthocentre of the excentral ΔI1I2I3.

16. Distance between special points:

(i) Distance between the circumcentre and orthocentre OH2 = R2(1-8 cos A cos B cos C)

(ii) Distance between the circumcentre and incentre OI2 = R2(1-8 sin (A/2) sin (B/2) sin (C/2) = R2 – 2Rr

(iii) Distance between the circumcentre and centroid OG2 = R219(a2+b2+c2)R^{2}-\frac{1}{9}\left ( a^{2}+b^{2} +c^{2}\right )

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