CBSE Class 11 Mathematics Solution of Triangles Notes Set 01

Relation between Sides and Angles

A triangle has six components, three sides and three angles. The three angles of a \( \triangle ABC \) are denoted by letters A, B, C and the sides opposite to these angles by letters a, b and c respectively. Following are some well known relations for a triangle (say \( \triangle ABC \)).
1. A + B + C = 180° (or π)
2. a + b > c, b + c > a, c + a > b
3. |a − b| < c, |b − c| < a, |c − a| < b

Generally, the relations involving the sides and angles of a triangle are cyclic in nature, e.g. to obtain the second similar relation to a + b > c, we simply replace a by b, b by c and c by a. So, to write all the relations, follow the cycles given.

The Law of Sines or Sine Rule

The sides of a triangle are proportional to the sines of the angles opposite to them, i.e.,

\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \text{ (say)} \]

More generally, if R be the radius of the circumcircle of the triangle ABC,

\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]

Note: 1. The above rule may also be expressed as \( \dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c} \).
2. The sine rule is very useful tool to express sides of a triangle in terms of sines of angle and vice-versa in the following manner:
\( \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = K \) (Let) \( \Rightarrow a = K\sin A,\ b = K\sin B,\ c = K\sin C \).
Similarly, \( \dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c} = \lambda \) (Let) \( \Rightarrow \sin A = \lambda a,\ \sin B = \lambda b,\ \sin C = \lambda c \).

Example 1. If the angles of a triangle are in the ratio 4 : 1 : 1, then the ratio of the longest side to the perimeter is [IIT Screening 2003]
(A) \( \sqrt{3} : (2 + \sqrt{3}) \)   (B) 1 : 6   (C) \( 1 : (2 + \sqrt{3}) \)   (D) 2 : 3
Answer:
Ans. (A)
4x + x + x = 180° ⇒ 6x = 180° ⇒ x = 30°
\( \dfrac{\sin 120°}{a} = \dfrac{\sin 30°}{b} = \dfrac{\sin 30°}{c} \)
∴ a : (a + b + c) = (sin 120°) : (sin 120° + sin 30° + sin 30°) = \( \dfrac{\sqrt{3}}{2} : \dfrac{\sqrt{3} + 2}{2} = \sqrt{3} : (\sqrt{3} + 2) \).

Example 2. In a triangle ABC, \( \angle B = \dfrac{\pi}{3} \) and \( \angle C = \dfrac{\pi}{4} \) and D divides BC internally in the ratio 1 : 3. Then \( \dfrac{\sin\angle BAD}{\sin\angle CAD} \) is equal to [UPSEAT 2003, 2001; IIT 1995]
(A) \( \dfrac{1}{3} \)   (B) \( \dfrac{1}{\sqrt{3}} \)   (C) \( \dfrac{1}{\sqrt{6}} \)   (D) \( \sqrt{\dfrac{2}{3}} \)
Answer:
Ans. (C)
Let \( \angle BAD = \alpha \), \( \angle CAD = \beta \) and BD = x, DC = 3x.
In \( \triangle ADB \), applying sine formulae, we get \( \dfrac{x}{\sin\alpha} = \dfrac{AD}{\sin(\pi/3)} \) ........(i)
In \( \triangle ADC \), applying sine formulae, we get \( \dfrac{3x}{\sin\beta} = \dfrac{AD}{\sin(\pi/4)} \) ..........(ii)
Dividing (i) by (ii), we get
\( \Rightarrow \dfrac{x}{\sin\alpha} \times \dfrac{\sin\beta}{3x} = \dfrac{AD}{\sin(\pi/3)} \times \dfrac{\sin(\pi/4)}{AD} \Rightarrow \dfrac{\sin\beta}{3\sin\alpha} = \dfrac{1/\sqrt{2}}{\sqrt{3}/2} = \sqrt{\dfrac{2}{3}} \Rightarrow \dfrac{\sin\beta}{\sin\alpha} = 3\sqrt{\dfrac{2}{3}} = \sqrt{6} \)
∴ \( \dfrac{\sin\angle BAD}{\sin\angle CAD} = \dfrac{\sin\alpha}{\sin\beta} = \dfrac{1}{\sqrt{6}} \).

Example 3. In a \( \triangle ABC \), A : B : C = 3 : 5 : 4. Then \( [a + b + c\sqrt{2}] \) is equal to [DCE 2001]
(A) 2b   (B) 2c   (C) 3b   (D) 3a
Answer:
Ans. (C)
A : B : C = 3 : 5 : 4 \( \Rightarrow \) A + B + C = 12x = 180° \( \Rightarrow \) x = 15°
∴ A = 45°, B = 75°, C = 60°
\( \dfrac{a}{\sin 45°} = \dfrac{b}{\sin 75°} = \dfrac{c}{\sin 60°} = k \) (say)
i.e., \( a = \dfrac{1}{\sqrt{2}}K,\ b = \dfrac{\sqrt{3}+1}{2\sqrt{2}}K,\ c = \dfrac{\sqrt{3}}{2}K \). Hence \( [a + b + c\sqrt{2}] = 3b \).

Example 4. In any triangle ABC if \( 2\cos B = \dfrac{a}{c} \), then the triangle is
(A) Right angled   (B) Equilateral   (C) Isosceles   (D) None of these
Answer:
Ans. (C)
\( 2\cos B = \dfrac{a}{c} = \dfrac{k\sin A}{k\sin C} = \dfrac{\sin A}{\sin C} \Rightarrow 2\cos B\sin C = \sin A \Rightarrow \sin(B + C) - \sin(B - C) = \sin A \)
\( \Rightarrow \sin(180° - A) - \sin(B - C) = \sin A \Rightarrow \sin A - \sin(B - C) = \sin A \Rightarrow \sin(B - C) = 0 \Rightarrow B - C = 0 \Rightarrow B = C \)
∴ Triangle is isosceles.

The Law of Cosines or Cosine Rule

In any triangle ABC, the square of any side is equal to the sum of the squares of the other two sides diminished by twice the product of these sides and the cosine of their included angle, that is for a triangle ABC,

(1) \( a^2 = b^2 + c^2 - 2bc\cos A \Rightarrow \cos A = \dfrac{b^2 + c^2 - a^2}{2bc} \)
(2) \( b^2 = c^2 + a^2 - 2ca\cos B \Rightarrow \cos B = \dfrac{c^2 + a^2 - b^2}{2ca} \)
(3) \( c^2 = a^2 + b^2 - 2ab\cos C \Rightarrow \cos C = \dfrac{a^2 + b^2 - c^2}{2ab} \)

Combining with \( \sin A = \dfrac{a}{2R},\ \sin B = \dfrac{b}{2R},\ \sin C = \dfrac{c}{2R} \) we have by division,
\( \tan A = \dfrac{abc}{R(b^2 + c^2 - a^2)},\quad \tan B = \dfrac{abc}{R(c^2 + a^2 - b^2)},\quad \tan C = \dfrac{abc}{R(a^2 + b^2 - c^2)} \)
where R is the radius of the circum-circle of the triangle ABC.

Example 5. The smallest angle of the \( \triangle ABC \), when \( a = 7,\ b = 4\sqrt{3} \) and \( c = \sqrt{13} \), is [MP PET 2003]
(A) 30°   (B) 15°   (C) 45°   (D) None of these
Answer:
Ans. (A)
Smallest angle is opposite to smaller side.
∴ \( \cos C = \dfrac{a^2 + b^2 - c^2}{2ab} = \dfrac{49 + 48 - 13}{2 \times 7 \times 4\sqrt{3}} = \dfrac{3}{2\sqrt{3}} = \dfrac{\sqrt{3}}{2} \Rightarrow \angle C = 30° \).

Example 6. In a \( \triangle ABC \), if \( \dfrac{b + c}{11} = \dfrac{c + a}{12} = \dfrac{a + b}{13} \), then \( \cos C = \) [Karnataka CET 2003]
(A) \( \dfrac{7}{5} \)   (B) \( \dfrac{5}{7} \)   (C) \( \dfrac{17}{36} \)   (D) \( \dfrac{16}{17} \)
Answer:
Ans. (B)
\( \dfrac{b + c}{11} = \dfrac{c + a}{12} = \dfrac{a + b}{13} = \lambda \) (Let)
∴ b + c = 11λ ......(i)   c + a = 12λ .......(ii)   and a + b = 13λ .......(iii)
From (i) + (ii) + (iii), 2(a + b + c) = 36λ ∴ a + b + c = 18λ .......(iv)
Now subtract (i), (ii) and (iii) from (iv): a = 7λ, b = 6λ, c = 5λ.
Now \( \cos C = \dfrac{a^2 + b^2 - c^2}{2ab} = \dfrac{(7\lambda)^2 + (6\lambda)^2 - (5\lambda)^2}{2 \cdot 7\lambda \cdot 6\lambda} = \dfrac{49\lambda^2 + 36\lambda^2 - 25\lambda^2}{84\lambda^2} = \dfrac{60\lambda^2}{84\lambda^2} = \dfrac{5}{7} \).

Example 7. In a \( \triangle ABC \), \( 2ac\sin\left(\dfrac{A - B + C}{2}\right) \) is equal to [IIT Screening 2000]
(A) \( a^2 + b^2 - c^2 \)   (B) \( c^2 + a^2 - b^2 \)   (C) \( b^2 - c^2 - a^2 \)   (D) \( c^2 - a^2 - b^2 \)
Answer:
Ans. (B)
\( 2ac\sin\dfrac{A - B + C}{2} = 2ac\sin\dfrac{\pi - 2B}{2} = 2ac\cos B = 2ac\dfrac{c^2 + a^2 - b^2}{2ca} = c^2 + a^2 - b^2 \).

Example 8. In ambiguous case if a, b and A are given and if there are two possible values of third side, are \( c_1 \) and \( c_2 \), then [UPSEAT 1999]
(A) \( c_1 - c_2 = 2\sqrt{a^2 + b^2\sin^2 A} \)   (B) \( c_1 - c_2 = 2\sqrt{a^2 - b^2\sin^2 A} \)   (C) \( c_1 - c_2 = 4\sqrt{a^2 + b^2\sin^2 A} \)   (D) \( c_1 - c_2 = 3\sqrt{a^2 - b^2\sin^2 A} \)
Answer:
Ans. (B)
\( \cos A = \dfrac{b^2 + c^2 - a^2}{2bc} \) or \( c^2 - (2b\cos A)c + (b^2 - a^2) = 0 \)
Which is quadratic equation in c. Let there be two roots, \( c_1 \) and \( c_2 \) of above quadratic equation then \( c_1 + c_2 = 2b\cos A \) and \( c_1c_2 = b^2 - a^2 \).
∴ \( c_1 - c_2 = \sqrt{(c_1 + c_2)^2 - 4c_1c_2} = \sqrt{(2b\cos A)^2 - 4(b^2 - a^2)} = \sqrt{4a^2 - 4b^2(1 - \cos^2 A)} = 2\sqrt{a^2 - b^2\sin^2 A} \).

Example 9. In a \( \triangle ABC \), \( \dfrac{\sin A}{\sin C} = \dfrac{\sin(A - B)}{\sin(B - C)} \), then \( a^2, b^2, c^2 \) are in [Karnataka CET 1999]
(A) A.P.   (B) G.P.   (C) H.P.   (D) None of these
Answer:
Ans. (A)
\( \dfrac{\sin A}{\sin C} = \dfrac{\sin A\cos B - \cos A\sin B}{\sin B\cos C - \cos B\sin C} \Rightarrow \dfrac{a}{c} = \dfrac{a\cos B - b\cos A}{b\cos C - c\cos B} \) (Using sine formula)
\( \Rightarrow ab\cos C - ac\cos B = ac\cos B - bc\cos A \Rightarrow ab\cos C + bc\cos A = 2ac\cos B \)
\( \Rightarrow \dfrac{a^2 + b^2 - c^2}{2} + \dfrac{b^2 + c^2 - a^2}{2} = \dfrac{c^2 + a^2 - b^2}{1} \Rightarrow b^2 = c^2 + a^2 - b^2 \Rightarrow b^2 = \dfrac{c^2 + a^2}{2} \)
\( \Rightarrow a^2, b^2, c^2 \) are in A.P.

Example 10. In a triangle ABC, AD is altitude from A. Given b > c, \( \angle C = 23° \) and \( AD = \dfrac{abc}{b^2 - c^2} \), \( \angle B \) is equal to [IIT 1994]
(A) 67°   (B) 44°   (C) 113°   (D) None of these
Answer:
Ans. (C)
\( \cos B = \dfrac{a^2 + c^2 - b^2}{2ac} = \dfrac{a^2 - (b^2 - c^2)}{2ac} \)
Now, \( AD = \dfrac{abc}{b^2 - c^2} \); ∴ \( \cos B = \dfrac{a^2 - \dfrac{abc}{AD}}{2ac} \)
Also \( AD = b\sin 23° \); ∴ \( \cos B = \dfrac{a - \dfrac{c}{\sin 23°}}{2c} \)
By sine formulae \( \dfrac{a}{c} = \dfrac{\sin(B + 23°)}{\sin 23°} \); ∴ \( \cos B = \dfrac{1}{2}\left[\dfrac{\sin(B + 23°)}{\sin 23°} - \dfrac{1}{\sin 23°}\right] \)
\( \Rightarrow \sin(B + 23°) - 2\sin 23°\cos B = 1 \Rightarrow \sin(23° - B) = -1 = \sin(-90°) \); therefore 23° − B = −90° or B = 113°.

Projection Formulae

In any triangle ABC, \( b\cos C + c\cos B = k\sin B\cos C + k\sin C\cos B \) (from sine rule)
\( = k\sin(B + C) = k\sin(\pi - A) = k\sin A = a \)

Similarly, we can deduce other projection formulae from sine rule.
(i) \( a = b\cos C + c\cos B \)    (ii) \( b = c\cos A + a\cos C \)    (iii) \( c = a\cos B + b\cos A \)
i.e., any side of a triangle is equal to the sum of the projection of other two sides on it.

Example 11. In a \( \triangle ABC \), \( \dfrac{\cos C + \cos A}{c + a} + \dfrac{\cos B}{b} \) is equal to [EAMCET 2001]
(A) \( \dfrac{1}{a} \)   (B) \( \dfrac{1}{b} \)   (C) \( \dfrac{1}{c} \)   (D) \( \dfrac{c + a}{b} \)
Answer:
Ans. (B)
\( \dfrac{\cos C + \cos A}{c + a} + \dfrac{\cos B}{b} = \dfrac{b(\cos C + \cos A) + (c + a)\cos B}{b(c + a)} = \dfrac{(b\cos C + c\cos B) + (b\cos A + a\cos B)}{b(c + a)} \)
\( = \dfrac{a + c}{b(c + a)} \) (Using projection formulae) \( = \dfrac{1}{b} \).

Example 12. If k be the perimeter of the \( \triangle ABC \), then \( b\cos^2\dfrac{C}{2} + c\cos^2\dfrac{B}{2} \) is equal to
(A) k   (B) 2k   (C) \( \dfrac{k}{2} \)   (D) None of these
Answer:
Ans. (C)
\( b\cos^2\dfrac{C}{2} + c\cos^2\dfrac{B}{2} = \dfrac{b}{2}(1 + \cos C) + \dfrac{c}{2}(1 + \cos B) = \dfrac{b}{2} + \dfrac{c}{2} + \dfrac{1}{2}(b\cos C + c\cos B) = \dfrac{a + b + c}{2} = \dfrac{k}{2} \).

Example 13. \( (a + b + c)(\cos A + \cos B + \cos C) = \)
(A) \( \sum a\sin^2\dfrac{A}{2} \)   (B) \( \sum a\cos^2\dfrac{A}{2} \)   (C) \( 2\sum a\sin^2\dfrac{A}{2} \)   (D) \( 2\sum a\cos^2\dfrac{A}{2} \)
Answer:
Ans. (D)
\( (a + b + c)(\cos A + \cos B + \cos C) \) = 9 terms.
\( = \sum a\cos A + \sum(b\cos C + c\cos B) = a\cos A + b\cos B + c\cos C + (a + b + c) = \sum a(1 + \cos A) = 2\sum a\cos^2\dfrac{A}{2} \).

Theorem of the Medians: Apollonius Theorem

In every triangle the sum of the squares of any two sides is equal to twice the square on half the third side together with twice the square on the median that bisects the third side.
For any triangle ABC, \( b^2 + c^2 = 2(h^2 + m^2) = 2\{m^2 + (a/2)^2\} \) by use of cosine rule.

If Δ be right angled, the mid point of hypotenuse is equidistant from the three vertices so that DA = DB = DC.
∴ \( b^2 + c^2 = a^2 \) which is pythagoras theorem. This theorem is very useful for solving problems of height and distance.

Example 14. If AD, BE and CF are the medians of a \( \triangle ABC \) then \( (AD^2 + BE^2 + CF^2) : (BC^2 + CA^2 + AB^2) \) is equal to
(A) 4 : 2   (B) 3 : 2   (C) 3 : 4   (D) 2 : 3
Answer:
Ans. (C)
We have, \( AB^2 + AC^2 = 2(AD^2 + BD^2) \Rightarrow \dfrac{c^2 + b^2}{2} - \dfrac{a^2}{4} = AD^2 \) ......(i)
\( \dfrac{a^2 + c^2}{2} - \dfrac{b^2}{4} = BE^2 \) ..........(ii) and \( \dfrac{a^2 + b^2}{2} - \dfrac{c^2}{4} = CF^2 \) ..........(iii)
\( a^2 + b^2 + c^2 - \dfrac{a^2 + b^2 + c^2}{4} = AD^2 + BE^2 + CF^2 \)
Adding (i), (ii) and (iii) we get, \( (AD^2 + BE^2 + CF^2) : (a^2 + b^2 + c^2) = 3 : 4 \).

Example 15. AD is a median of the \( \triangle ABC \), if AE and AF are medians of the triangles ABD and ADC respectively and \( AD = m_1,\ AE = m_2,\ AF = m_3 \), then \( \dfrac{a^2}{8} \) is equal to
(A) \( m_2^2 + m_3^2 - 2m_1^2 \)   (B) \( m_1^2 + m_2^2 - 2m_3^2 \)   (C) \( m_2^2 + m_3^2 - m_1^2 \)   (D) None of these
Answer:
Ans. (A)
In \( \triangle ABC \), \( AD^2 = m_1^2 = \dfrac{c^2 + b^2}{2} - \dfrac{a^2}{4} \)
In \( \triangle ABD \), \( AE^2 = m_2^2 = \dfrac{c^2 + AD^2}{2} - \dfrac{(a/2)^2}{4} \)
In \( \triangle ADC \), \( AF^2 = m_3^2 = \dfrac{AD^2 + b^2}{2} - \dfrac{(a/2)^2}{4} \)
∴ \( m_2^2 + m_3^2 = AD^2 + \dfrac{b^2 + c^2}{2} - \dfrac{a^2}{8} = m_1^2 + m_1^2 + \dfrac{a^2}{4} - \dfrac{a^2}{8} \)
\( m_2^2 + m_3^2 = 2m_1^2 + \dfrac{a^2}{8} \Rightarrow \dfrac{a^2}{8} = m_2^2 + m_3^2 - 2m_1^2 \).

Napier's Analogy (Law of Tangents)

For any triangle ABC,

(1) \( \tan\left(\dfrac{A - B}{2}\right) = \left(\dfrac{a - b}{a + b}\right)\cot\dfrac{C}{2} \)
(2) \( \tan\left(\dfrac{B - C}{2}\right) = \left(\dfrac{b - c}{b + c}\right)\cot\dfrac{A}{2} \)
(3) \( \tan\left(\dfrac{C - A}{2}\right) = \left(\dfrac{c - a}{c + a}\right)\cot\dfrac{B}{2} \)

Note: Mollweide's formula: For any triangle, \( \dfrac{a + b}{c} = \dfrac{\cos\frac{1}{2}(A - B)}{\sin\frac{1}{2}C},\quad \dfrac{a - b}{c} = \dfrac{\sin\frac{1}{2}(A - B)}{\cos\frac{1}{2}C} \).

Example 16. If \( \tan\dfrac{B - C}{2} = x\cot\dfrac{A}{2} \), then x equal to [MP PET 1992, 2002]
(A) \( \dfrac{c - a}{c + a} \)   (B) \( \dfrac{a - b}{a + b} \)   (C) \( \dfrac{b - c}{b + c} \)   (D) None of these
Answer:
Ans. (C)
We know, \( \tan\dfrac{B - C}{2} = \dfrac{b - c}{b + c}\cot\dfrac{A}{2} \Rightarrow x = \dfrac{b - c}{b + c} \).

Example 17. If in a \( \triangle ABC \), a = 6, b = 3 and \( \cos(A - B) = \dfrac{4}{5} \), then [Roorkee 1997]
(A) \( C = \dfrac{\pi}{4} \)   (B) \( A = \sin^{-1}\dfrac{2}{\sqrt{5}} \)   (C) \( ar(\triangle ABC) = 9 \)   (D) None of these
Answer:
Ans. (B, C)
\( \tan\dfrac{A - B}{2} = \dfrac{a - b}{a + b}\cot\dfrac{C}{2} = \dfrac{1}{3}\cot\dfrac{C}{2} \)
∴ \( \dfrac{4}{5} = \cos(A - B) = \dfrac{1 - \tan^2\frac{A - B}{2}}{1 + \tan^2\frac{A - B}{2}} = \dfrac{1 - \frac{1}{9}\cot^2\frac{C}{2}}{1 + \frac{1}{9}\cot^2\frac{C}{2}} \)
∴ \( \tan^2\dfrac{C}{2} = 1 \Rightarrow C = \dfrac{\pi}{2} \) ∴ \( ar(\triangle ABC) = \dfrac{1}{2}ab = \dfrac{1}{2} \cdot 6 \cdot 3 = 9 \)
Also, \( \sin A = \dfrac{6}{\sqrt{3^2 + 6^2}} = \dfrac{2}{\sqrt{5}} \).

Example 18. In \( \triangle ABC \), \( A = \dfrac{\pi}{3} \) and b : c = 2 : 3. If \( \tan\alpha = \dfrac{\sqrt{3}}{5} \), \( 0 < \alpha < \dfrac{\pi}{2} \), then
(A) \( B = 60° + \alpha \)   (B) \( C = 60° + \alpha \)   (C) \( B = 60° - \alpha \)   (D) \( C = 60° - \alpha \)
Answer:
Ans. (B, C)
\( \tan\dfrac{C - B}{2} = \dfrac{c - b}{c + b}\cot\dfrac{A}{2} \Rightarrow \tan\dfrac{C - B}{2} = \dfrac{1}{5}\cot 30° = \dfrac{\sqrt{3}}{5} = \tan\alpha \)
∴ C − B = 2α and C + B = 180° − 60° = 120°, i.e., B = 60° − α, C = 60° + α.

Example 19. In a \( \triangle ABC \), a = 2b and \( |A - B| = \dfrac{\pi}{3} \). The measure of \( \angle C \) is
(A) \( \dfrac{\pi}{4} \)   (B) \( \dfrac{\pi}{3} \)   (C) \( \dfrac{\pi}{6} \)   (D) None of these
Answer:
Ans. (B)
Clearly, A > B (∵ a > b)
Now \( \tan\dfrac{A - B}{2} = \dfrac{a - b}{a + b}\cot\dfrac{C}{2} \Rightarrow \tan 30° = \dfrac{1}{3}\cot\dfrac{C}{2} \)
∴ \( \sqrt{3} = \cot\dfrac{C}{2} \) or \( \dfrac{C}{2} = \dfrac{\pi}{6} \Rightarrow C = \dfrac{\pi}{3} \).

Area of Triangle

Let three angles of \( \triangle ABC \) are denoted by A, B, C and the sides opposite to these angles by letters a, b, c respectively.

(1) When two sides and the included angle be given: The area of triangle ABC is given by,
\( \Delta = \dfrac{1}{2}bc\sin A = \dfrac{1}{2}ca\sin B = \dfrac{1}{2}ab\sin C \)   i.e., \( \Delta = \dfrac{1}{2} \) (Product of two sides) × sine of included angle

(2) When three sides are given: Area of \( \triangle ABC = \Delta = \sqrt{s(s - a)(s - b)(s - c)} \)
where semiperimeter of triangle \( s = \dfrac{a + b + c}{2} \)

(3) When three sides and the circum-radius be given: Area of triangle \( \Delta = \dfrac{abc}{4R} \), where R be the circum-radius of the triangle.

(4) When two angles and included side be given:
\( \Delta = \dfrac{1}{2}a^2\dfrac{\sin B\sin C}{\sin(B + C)} = \dfrac{1}{2}b^2\dfrac{\sin A\sin C}{\sin(A + C)} = \dfrac{1}{2}c^2\dfrac{\sin A\sin B}{\sin(A + B)} \)

Example 20. In a \( \triangle ABC \) if a = 2x, b = 2y and \( \angle C = 120° \), then the area of the triangle is [MP PET 1986, 2002]
(A) xy   (B) \( xy\sqrt{3} \)   (C) 3xy   (D) 2xy
Answer:
Ans. (B)
\( \Delta = \dfrac{1}{2}ab\sin C = \dfrac{1}{2} \cdot 2x \cdot 2y \sin 120° = \sqrt{3}\,xy \).

Example 21. In a \( \triangle ABC \), \( \dfrac{\cos A}{a} = \dfrac{\cos B}{b} = \dfrac{\cos C}{c} \) and a = 2, then the area of a triangle is [MP PET 2000; IIT 1993]
(A) 1   (B) 2   (C) \( \dfrac{\sqrt{3}}{2} \)   (D) \( \sqrt{3} \)
Answer:
Ans. (D)
By sine rule, tan A = tan B = tan C ∴ Triangle is equilateral.
Hence, \( \Delta = \dfrac{1}{2} \cdot a \cdot a \cdot \sin 60° = \dfrac{1}{2} \cdot 2 \cdot 2 \cdot \dfrac{\sqrt{3}}{2} = \sqrt{3} \).

Example 22. In a triangle ABC, a, b, A are given and \( c_1, c_2 \) are two values of third side c. The sum of the areas of triangles with sides a, b, \( c_1 \) and a, b, \( c_2 \) is
(A) \( \dfrac{1}{2}a^2\sin 2A \)   (B) \( \dfrac{1}{2}b^2\sin 2A \)   (C) \( b^2\sin 2A \)   (D) \( a^2\sin 2A \)
Answer:
Ans. (B)
Let the triangles be \( \Delta_1 = ABC_1 \) and \( \Delta_2 = ABC_2 \). A, b, a are given and c has two values. Hence we apply cosine formulae
\( \cos A = \dfrac{b^2 + c^2 - a^2}{2bc} \) or \( c^2 - 2bc\cos A + b^2 - a^2 = 0 \).
Above is quadratic in c. If \( c_1, c_2 \) be the two values of c, then \( c_1 + c_2 = 2b\cos A,\ c_1c_2 = b^2 - a^2 \)
\( \Delta_1 = \dfrac{1}{2}bc_1\sin A,\ \Delta_2 = \dfrac{1}{2}bc_2\sin A \)
∴ \( \Delta_1 + \Delta_2 = \dfrac{1}{2}b\sin A\,(c_1 + c_2) = \dfrac{1}{2}b\sin A\,(2b\cos A) = b^2\sin A\cos A = \dfrac{1}{2}b^2\sin 2A \).

Example 23. If Δ stands for the area of a triangle ABC, then \( a^2\sin 2B + b^2\sin 2A = \) [WB JEE 1988]
(A) 3Δ   (B) 2Δ   (C) 4Δ   (D) −4Δ
Answer:
Ans. (C)
Use sine rule, \( \Delta = \dfrac{1}{2}ab\sin C \)
L.H.S. = \( k^2(\sin^2 A \cdot 2\sin B\cos B + \sin^2 B \cdot 2\sin A\cos A) = k^2[2\sin A \cdot \sin B \cdot \sin(A + B)] = 2ab\sin C = 4\Delta \).

Example 24. In a \( \triangle ABC \), \( A = \dfrac{2\pi}{3},\ b - c = 3\sqrt{3} \) and \( ar(\triangle ABC) = \dfrac{9\sqrt{3}}{2}\ \text{cm}^2 \). Then a is
(A) \( 6\sqrt{3} \) cm   (B) 9 cm   (C) 18 cm   (D) None of these
Answer:
Ans. (B)
\( \dfrac{1}{2}bc\sin\dfrac{2\pi}{3} = \dfrac{9\sqrt{3}}{2} \) or \( \dfrac{1}{2} \cdot \dfrac{\sqrt{3}}{2} \cdot bc = \dfrac{9\sqrt{3}}{2} \Rightarrow bc = 18 \)
Also, \( \cos\dfrac{2\pi}{3} = \dfrac{b^2 + c^2 - a^2}{2bc} \Rightarrow -\dfrac{1}{2} = \dfrac{(b - c)^2 + 2bc - a^2}{2bc} \) or \( (b - c)^2 + 3bc - a^2 = 0 \) or 27 + 54 = \( a^2 \) ⇒ a = 9.

Example 25. If \( p_1, p_2, p_3 \) are altitudes of a triangle ABC from the vertices A, B, C and Δ the area of the triangle, then \( p_1^{-2} + p_2^{-2} + p_3^{-2} \) is equal to
(A) \( \dfrac{a + b + c}{\Delta} \)   (B) \( \dfrac{a^2 + b^2 + c^2}{4\Delta^2} \)   (C) \( \dfrac{a^2 + b^2 + c^2}{\Delta^2} \)   (D) None of these
Answer:
Ans. (B)
We have \( \dfrac{1}{2}ap_1 = \dfrac{1}{2}bp_2 = \dfrac{1}{2}cp_3 = \Delta \Rightarrow p_1 = \dfrac{2\Delta}{a},\ p_2 = \dfrac{2\Delta}{b},\ p_3 = \dfrac{2\Delta}{c} \)
∴ \( \dfrac{1}{p_1^2} + \dfrac{1}{p_2^2} + \dfrac{1}{p_3^2} = \dfrac{a^2 + b^2 + c^2}{4\Delta^2} \).

Half Angle Formulae

If 2s shows the perimeter of a triangle ABC, i.e., 2s = a + b + c, then

(1) Formulae for \( \sin\dfrac{A}{2},\ \sin\dfrac{B}{2},\ \sin\dfrac{C}{2} \):
(i) \( \sin\dfrac{A}{2} = \sqrt{\dfrac{(s - b)(s - c)}{bc}} \)   (ii) \( \sin\dfrac{B}{2} = \sqrt{\dfrac{(s - a)(s - c)}{ca}} \)   (iii) \( \sin\dfrac{C}{2} = \sqrt{\dfrac{(s - a)(s - b)}{ab}} \)

(2) Formulae for \( \cos\dfrac{A}{2},\ \cos\dfrac{B}{2},\ \cos\dfrac{C}{2} \):
(i) \( \cos\dfrac{A}{2} = \sqrt{\dfrac{s(s - a)}{bc}} \)   (ii) \( \cos\dfrac{B}{2} = \sqrt{\dfrac{s(s - b)}{ca}} \)   (iii) \( \cos\dfrac{C}{2} = \sqrt{\dfrac{s(s - c)}{ab}} \)

(3) Formulae for \( \tan\dfrac{A}{2},\ \tan\dfrac{B}{2},\ \tan\dfrac{C}{2} \):
(i) \( \tan\dfrac{A}{2} = \sqrt{\dfrac{(s - b)(s - c)}{s(s - a)}} \)   (ii) \( \tan\dfrac{B}{2} = \sqrt{\dfrac{(s - c)(s - a)}{s(s - b)}} \)   (iii) \( \tan\dfrac{C}{2} = \sqrt{\dfrac{(s - a)(s - b)}{s(s - c)}} \)

Note: 1. \( \sin A = 2\sin\dfrac{A}{2}\cos\dfrac{A}{2} = 2\sqrt{\dfrac{(s - b)(s - c)}{bc}}\sqrt{\dfrac{s(s - a)}{bc}} = \dfrac{2}{bc}\sqrt{s(s - a)(s - b)(s - c)} = \dfrac{2\Delta}{bc} \)
Similarly \( \sin B = \dfrac{2\Delta}{ca},\ \sin C = \dfrac{2\Delta}{ab} \).
2. \( \tan\dfrac{A}{2} = \dfrac{(s - b)(s - c)}{\Delta},\ \tan\dfrac{B}{2} = \dfrac{(s - c)(s - a)}{\Delta},\ \tan\dfrac{C}{2} = \dfrac{(s - a)(s - b)}{\Delta} \)

Important Tips

1. \( \tan\dfrac{A}{2}\tan\dfrac{B}{2} = \left\{\dfrac{(s - b)(s - c)}{s(s - a)} \cdot \dfrac{(s - c)(s - a)}{s(s - b)}\right\}^{1/2} = \dfrac{s - c}{s} \Rightarrow \cot\dfrac{A}{2}\cot\dfrac{B}{2} = \dfrac{s}{s - c} \)

2. \( \tan\dfrac{A}{2} + \tan\dfrac{B}{2} = \sqrt{\dfrac{s - c}{s}}\left[\sqrt{\dfrac{s - b}{s - a}} + \sqrt{\dfrac{s - a}{s - b}}\right] = \sqrt{\dfrac{s - c}{s}}\left[\dfrac{s - b + s - a}{\sqrt{(s - a)(s - b)}}\right] = \dfrac{c}{s}\left[\dfrac{s(s - c)}{(s - a)(s - b)}\right]^{1/2} = \dfrac{c}{s}\cot\dfrac{C}{2} \)
Another form: \( \dfrac{c \cdot (s - c)}{\{s(s - a)(s - b)(s - c)\}^{1/2}} = \dfrac{c}{\Delta}(s - c) \)

3. \( \tan\dfrac{A}{2} - \tan\dfrac{B}{2} = \dfrac{a - b}{\Delta}(s - c) \)

4. \( \cot\dfrac{A}{2} + \cot\dfrac{B}{2} = \dfrac{\tan\frac{A}{2} + \tan\frac{B}{2}}{\tan\frac{A}{2}\tan\frac{B}{2}} = \dfrac{c}{s - c}\cot\dfrac{C}{2} \)

Example 26. If in any \( \triangle ABC \); \( \cot\dfrac{A}{2},\ \cot\dfrac{B}{2},\ \cot\dfrac{C}{2} \) are in A.P., then [MP PET 2003]
(A) \( \cot\dfrac{A}{2}\cot\dfrac{B}{2} = 4 \)   (B) \( \cot\dfrac{A}{2}\cot\dfrac{C}{2} = 3 \)   (C) \( \cot\dfrac{B}{2}\cot\dfrac{C}{2} = 1 \)   (D) \( \cot\dfrac{B}{2}\tan\dfrac{C}{2} = 0 \)
Answer:
Ans. (B)
Trick: Take A = B = C = 60°, then \( \cot\dfrac{A}{2},\ \cot\dfrac{B}{2} \) and \( \cot\dfrac{C}{2} \) are in A.P. with common difference zero. Now option (B) satisfies.

Example 27. In a \( \triangle ABC \), if 3a = b + c, then the value of \( \cot\dfrac{B}{2}\cot\dfrac{C}{2} \) is [Roorkee 1986; MP PET 1990, 97, 98]
(A) 1   (B) 2   (C) \( \sqrt{3} \)   (D) \( \sqrt{2} \)
Answer:
Ans. (B)
\( \cot\dfrac{B}{2}\cot\dfrac{C}{2} = \sqrt{\dfrac{s(s - b)}{(s - a)(s - c)}}\sqrt{\dfrac{s(s - c)}{(s - a)(s - b)}} = \dfrac{s}{s - a} \)
Given 3a = b + c \( \Rightarrow \) a + b + c = 4a ∴ \( \cot\dfrac{B}{2}\cot\dfrac{C}{2} = \dfrac{2a}{a} = 2 \).

Example 28. In a triangle ABC, \( \tan\dfrac{A}{2} = \dfrac{5}{6} \) and \( \tan\dfrac{C}{2} = \dfrac{2}{5} \), then [EAMCET 1994]
(A) a, b, c are in A.P.   (B) cos A, cos B, cos C are in A.P.   (C) sin A, sin B, sin C are in A.P.   (D) Both (A) and (C)
Answer:
Ans. (D)
\( \tan\dfrac{A}{2}\tan\dfrac{C}{2} = \dfrac{s - b}{s} \Rightarrow \dfrac{5}{6} \cdot \dfrac{2}{5} = \dfrac{s - b}{s} \Rightarrow 2s = 3b \Rightarrow a + b + c = 3b \Rightarrow a + c = 2b \).

Example 29. If the sides of triangle a, b, c are in A.P. then \( \tan\dfrac{A}{2} + \tan\dfrac{C}{2} \) is equal to [Roorkee 1993]
(A) \( \dfrac{2}{3}\cot\dfrac{A}{2} \)   (B) \( \dfrac{2}{3}\cot\dfrac{B}{2} \)   (C) \( \dfrac{2}{3}\cot\dfrac{C}{2} \)   (D) None of these
Answer:
Ans. (B)
\( \tan\dfrac{A}{2} + \tan\dfrac{C}{2} = \sqrt{\dfrac{(s - b)(s - c)}{s(s - a)}} + \sqrt{\dfrac{(s - a)(s - b)}{s(s - c)}} = \dfrac{b}{s}\cot\dfrac{B}{2} = \dfrac{2b}{2s}\cot\dfrac{B}{2} \)
∵ a, b, c in A.P. ∴ a + c = 2b ⇒ 2s = 3b
Hence, \( \tan\dfrac{A}{2} + \tan\dfrac{C}{2} = \dfrac{2b}{3b}\cot\dfrac{B}{2} = \dfrac{2}{3}\cot\dfrac{B}{2} \).

Example 30. In \( \triangle ABC \), \( \left(\cot\dfrac{A}{2} + \cot\dfrac{B}{2}\right)\left(a\sin^2\dfrac{B}{2} + b\sin^2\dfrac{A}{2}\right) \) equal to [Roorkee 1988]
(A) cot C   (B) c cot C   (C) \( \cot\dfrac{C}{2} \)   (D) \( c\cot\dfrac{C}{2} \)
Answer:
Ans. (D)
\( \left(\cot\dfrac{A}{2} + \cot\dfrac{B}{2}\right)\left(a\sin^2\dfrac{B}{2} + b\sin^2\dfrac{A}{2}\right) = \dfrac{\cos\frac{C}{2}}{\sin\frac{A}{2}\sin\frac{B}{2}}\left(a\sin^2\dfrac{B}{2} + b\sin^2\dfrac{A}{2}\right) = \cos\dfrac{C}{2}\left(a\dfrac{\sin\frac{B}{2}}{\sin\frac{A}{2}} + b\dfrac{\sin\frac{A}{2}}{\sin\frac{B}{2}}\right) \)
\( = \sqrt{\dfrac{s(s - c)}{ab}}\left\{\sqrt{ab}\sqrt{\dfrac{s - a}{s - b}} + \sqrt{ab}\sqrt{\dfrac{s - b}{s - a}}\right\} = \sqrt{s(s - c)}\,\dfrac{(s - a) + (s - b)}{\sqrt{(s - a)(s - b)}} = c\sqrt{\dfrac{s(s - c)}{(s - a)(s - b)}} = c\cot\dfrac{C}{2} \).
Trick: Such type of unconditional problems can be checked by putting the particular values for a = 1, \( b = \sqrt{3} \), c = 2 and A = 30°, B = 60°, C = 90°. Here expression is equal to 2 which is given by (D).