JEE Mathematics Properties and Solution of Triangles Notes Set 01

Sine Rule and Cosine Rule

In any triangle ABC with the usual notation, \( \frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}=2R \), where R is the circumradius. Hence \( a=2R\sin A \), and every side can be replaced by the sine of the opposite angle.
Cosine rule: \( \cos A=\frac{b^2+c^2-a^2}{2bc} \), and cyclically for B and C. It is used when two sides and the included angle, or all three sides, are known.
Projection rule: \( a=b\cos C+c\cos B \), \( b=c\cos A+a\cos C \), \( c=a\cos B+b\cos A \).
Ambiguous case: given a, b and the angle A, the sine rule gives \( \sin B=\frac{b\sin A}{a} \). If this exceeds 1 no triangle exists; if it is less than 1 there may be one or two triangles, depending on whether a is shorter than b and whether a is at least \( b\sin A \). In the two-triangle case the two values of the third side add up to \( 2b\cos A \).

Napier's Analogy and Half-Angle Formulae

Tangent rule: \( \tan\frac{B-C}2=\frac{b-c}{b+c}\cot\frac A2 \), and cyclically. It gives the two unknown angles when two sides and the included angle are known.
With \( s=\frac{a+b+c}2 \): \( \sin\frac A2=\sqrt{\frac{(s-b)(s-c)}{bc}} \), \( \cos\frac A2=\sqrt{\frac{s(s-a)}{bc}} \), \( \tan\frac A2=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \).
Useful product: \( \cot\frac B2\cot\frac C2=\frac{s}{s-a} \).

Area of a Triangle

\( \Delta=\frac12bc\sin A=\frac12ca\sin B=\frac12ab\sin C \).
Heron's formula: \( \Delta=\sqrt{s(s-a)(s-b)(s-c)} \).
\( \Delta=\frac{abc}{4R}=rs=2R^2\sin A\sin B\sin C \).

Circumradius, Inradius and Exradii

Circumradius: \( R=\frac{abc}{4\Delta}=\frac a{2\sin A} \).
Inradius: \( r=\frac\Delta s=(s-a)\tan\frac A2=4R\sin\frac A2\sin\frac B2\sin\frac C2 \).
Exradius opposite A: \( r_1=\frac\Delta{s-a}=s\tan\frac A2=4R\sin\frac A2\cos\frac B2\cos\frac C2 \); similarly \( r_2=\frac\Delta{s-b} \) and \( r_3=\frac\Delta{s-c} \).
Relations: \( \frac1r=\frac1{r_1}+\frac1{r_2}+\frac1{r_3} \), \( r_1+r_2+r_3=4R+r \), \( r_1r_2+r_2r_3+r_3r_1=s^2 \), \( rr_1r_2r_3=\Delta^2 \), \( r_1-r=\frac{\Delta a}{s(s-a)}=a\tan\frac A2 \).

Medians, Bisectors and Altitudes

Median from A: \( m_a=\frac12\sqrt{2b^2+2c^2-a^2} \); the sum of the squares of the medians is \( \frac34(a^2+b^2+c^2) \).
Internal bisector of A: \( \ell_a=\frac{2bc\cos\frac A2}{b+c} \). Altitude on side a: \( \frac{2\Delta}{a} \).

Special Points and Triangles

Orthocentre H: the circumradius of each of the triangles HBC, HCA, HAB equals R. Distance of the circumcentre from side a is \( R\cos A \); distance of the incentre from vertex A is \( r\,\mathrm{cosec}\frac A2 \).
Pedal triangle (feet of the altitudes): sides \( a\cos A,\ b\cos B,\ c\cos C \), angles \( \pi-2A,\ \pi-2B,\ \pi-2C \), circumradius \( \frac R2 \) (for an acute triangle).
Excentral triangle I1I2I3: sides \( 4R\cos\frac A2 \) and so on, angles \( \frac\pi2-\frac A2 \), circumradius 2R; the incentre I is its orthocentre and ABC is its pedal triangle.

Worked Examples

Worked Example 1. Can a triangle be constructed with a = 5, b = 7 and \( \sin A=\frac34 \)?
Answer:
By the sine rule \( \sin B=\frac{b\sin A}{a}=\frac{7\times3/4}{5}=\frac{21}{20}>1 \), which is impossible, so no such triangle exists.

Worked Example 2. If in a triangle ABC the angles are in A.P., show that \( a^2-ac+c^2=b^2 \).
Answer:
A + C = 2B gives 3B = 180°, so B = 60°.
Cosine rule: \( \cos60^\circ=\frac{a^2+c^2-b^2}{2ac}=\frac12 \), hence \( a^2+c^2-b^2=ac \).

Worked Example 3. Find the unknown elements of the triangle with \( a=\sqrt3+1,\ b=\sqrt3-1,\ C=90^\circ \).
Answer:
\( A+B=90^\circ \). Napier: \( \tan\frac{A-B}{2}=\frac{a-b}{a+b}\cot45^\circ=\frac{2}{2\sqrt3}=\frac1{\sqrt3} \), so \( A-B=60^\circ \).
Hence \( A=75^\circ,\ B=15^\circ \) and \( c=\sqrt{a^2+b^2}=\sqrt8=2\sqrt2 \).

Worked Example 4. In a triangle ABC prove that \( a(b\cos C-c\cos B)=b^2-c^2 \).
Answer:
\( b\cos C=\frac{a^2+b^2-c^2}{2a} \) and \( c\cos B=\frac{a^2+c^2-b^2}{2a} \).
Subtracting, \( b\cos C-c\cos B=\frac{2b^2-2c^2}{2a}=\frac{b^2-c^2}{a} \); multiply by a.