CBSE Class 11 Mathematics Solution of Triangles Notes Set 03

Regular Polygon

A regular polygon is a polygon which has all its sides equal and all its angles equal.
(1) Each interior angle of a regular polygon of n sides is \( \left(\dfrac{2n - 4}{n}\right) \) × right angles = \( \left(\dfrac{2n - 4}{n} \times \dfrac{\pi}{2}\right) \) radians.
(2) The circle passing through all the vertices of a regular polygon is called its circumscribed circle. If a is the length of each side of a regular polygon of n sides, then the radius R of the circumscribed circle, is given by \( R = \dfrac{a}{2}\csc\left(\dfrac{\pi}{n}\right) \)
(3) The circle which can be inscribed within the regular polygon so as to touch all its sides is called its inscribed circle. Again if a is the length of each side of a regular polygon of n sides, then the radius r of the inscribed circle is given by \( r = \dfrac{a}{2}\cot\left(\dfrac{\pi}{n}\right) \)

(4) The area of a regular polygon is given by Δ = n × area of triangle OAB
\( = \dfrac{1}{4}na^2\cot\left(\dfrac{\pi}{n}\right) \) (in terms of side)
\( = nr^2\tan\left(\dfrac{\pi}{n}\right) \) (in terms of in-radius)
\( = \dfrac{n}{2}R^2\sin\left(\dfrac{2\pi}{n}\right) \) (in terms of circum-radius)

Example 43. The sum of the radii of inscribed and circumscribed circles for an n sided regular polygon of side a, is [AIEEE 2003]
(A) \( a\cot\left(\dfrac{\pi}{n}\right) \)   (B) \( \dfrac{a}{2}\cot\left(\dfrac{\pi}{2n}\right) \)   (C) \( a\cot\left(\dfrac{\pi}{2n}\right) \)   (D) \( \dfrac{a}{4}\cot\left(\dfrac{\pi}{2n}\right) \)
Answer:
Ans. (B)
\( \tan\left(\dfrac{\pi}{n}\right) = \dfrac{a}{2r} \) and \( \sin\left(\dfrac{\pi}{n}\right) = \dfrac{a}{2R} \Rightarrow r + R = \dfrac{a}{2}\left[\cot\dfrac{\pi}{n} + \csc\dfrac{\pi}{n}\right] = \dfrac{a}{2}\cot\left(\dfrac{\pi}{2n}\right) \).

Example 44. The area of the circle and the area of a regular polygon of n sides and its perimeter equal to that of the circle are in the ratio of
(A) \( \tan\left(\dfrac{\pi}{n}\right) : \dfrac{\pi}{n} \)   (B) \( \cos\left(\dfrac{\pi}{n}\right) : \dfrac{\pi}{n} \)   (C) \( \sin\left(\dfrac{\pi}{n}\right) : \dfrac{\pi}{n} \)   (D) \( \cot\left(\dfrac{\pi}{n}\right) : \dfrac{\pi}{n} \)
Answer:
Ans. (A)
Let r be the radius of the circle and \( A_1 \) be its area ∴ \( A_1 = \pi r^2 \)
Since the perimeter of the circle is the same as the perimeter of a regular polygon of n sides ∴ 2πr = na, where 'a' is the length of one side of the regular polygon, ∴ \( a = \dfrac{2\pi r}{n} \)
Let \( A_2 \) be the area of the polygon, then \( A_2 = \dfrac{1}{4}na^2\cot\dfrac{\pi}{n} = \dfrac{1}{4}n \cdot \dfrac{4\pi^2r^2}{n^2}\cot\dfrac{\pi}{n} = \pi r^2 \cdot \dfrac{\pi}{n} \cdot \cot\dfrac{\pi}{n} \)
∴ \( A_1 : A_2 = \pi r^2 : \pi r^2 \cdot \dfrac{\pi}{n}\cot\dfrac{\pi}{n} = 1 : \dfrac{\pi}{n}\cot\dfrac{\pi}{n} = \tan\dfrac{\pi}{n} : \dfrac{\pi}{n} \).

Example 45. A regular polygon of nine sides, each of length 2 is inscribed in a circle. The radius of the circle is [IIT 1994]
(A) \( \csc\dfrac{\pi}{9} \)   (B) \( \csc\dfrac{\pi}{3} \)   (C) \( \cot\dfrac{\pi}{9} \)   (D) \( \tan\dfrac{\pi}{9} \)
Answer:
Ans. (A)
We know that radius of the circumcircle is given by \( R = \dfrac{a}{2}\csc\left(\dfrac{\pi}{n}\right) \) ; Here, a = 2, n = 9
∴ Required radius = \( \dfrac{2}{2}\csc\dfrac{\pi}{9} = \csc\dfrac{\pi}{9} \).

Example 46. If the number of sides of two regular polygons having the same perimeter be n and 2n respectively, their areas are in the ratio
(A) \( \dfrac{2\cos\frac{\pi}{n}}{\cos\frac{\pi}{2n}} \)   (B) \( \dfrac{2\cos\frac{\pi}{n}}{1 + \cos\frac{\pi}{n}} \)   (C) \( \dfrac{\cos\frac{\pi}{n}}{\sin\frac{\pi}{n}} \)   (D) None of these
Answer:
Ans. (B)
Let s be the perimeter of both the polygons. Then the length of each side of the first polygon is \( \dfrac{s}{n} \) and that of second polygon is \( \dfrac{s}{2n} \).
If \( A_1, A_2 \) denote their areas, then \( A_1 = \dfrac{n}{4}\left[\dfrac{s}{n}\right]^2\cot\dfrac{\pi}{n} \) and \( A_2 = \dfrac{1}{4}(2n)\left(\dfrac{s}{2n}\right)^2\cot\dfrac{\pi}{2n} \)
\( \dfrac{A_1}{A_2} = \dfrac{2\cot\frac{\pi}{n}}{\cot\frac{\pi}{2n}} = \dfrac{2\cos\frac{\pi}{n}\sin\frac{\pi}{2n}}{\sin\frac{\pi}{n}\cos\frac{\pi}{2n}} = \dfrac{2\cos\frac{\pi}{n}\sin\frac{\pi}{2n}}{2\sin\frac{\pi}{2n}\cos\frac{\pi}{2n}\cos\frac{\pi}{2n}} \Rightarrow \dfrac{A_1}{A_2} = \dfrac{2\cos\frac{\pi}{n}}{1 + \cos\frac{\pi}{n}} \).

Solutions of Triangles

Different formulae will be used in different cases and sometimes the same problem may be solved in different ways by different formulae. We should, therefore, look for that formula which will suit the problem best.
(1) Solution of a right angled triangle
(2) Solution of a triangle in general

(1) Solution of a Right Angled Triangle

(i) When two sides are given: Let the triangle be right angled at C. Then we can determine the remaining elements as given in the following table

(ii) When a side and an acute angle are given: In this case, we can determine the remaining elements as given in the following table

(2) Solution of a Triangle in General

(i) When three sides a, b and c are given: In this case, the remaining elements are determined by using the following formulae, \( \Delta = \sqrt{s(s - a)(s - b)(s - c)} \), where 2s = a + b + c = perimeter of triangle
\( \sin A = \dfrac{2\Delta}{bc},\quad \sin B = \dfrac{2\Delta}{ac},\quad \sin C = \dfrac{2\Delta}{ab} \)
\( \tan\dfrac{A}{2} = \dfrac{\Delta}{s(s - a)},\quad \tan\dfrac{B}{2} = \dfrac{\Delta}{s(s - b)},\quad \tan\dfrac{C}{2} = \dfrac{\Delta}{s(s - c)} \)

(ii) When two sides a, b and the included angle C are given: In this case, we use the following formulae
\( \Delta = \dfrac{1}{2}ab\sin C \) ;  \( \tan\dfrac{A - B}{2} = \dfrac{a - b}{a + b}\cot\dfrac{C}{2} \) ;  \( \dfrac{A + B}{2} = 90° - \dfrac{C}{2} \) and \( c = \dfrac{a\sin C}{\sin A} \)

(iii) When one side a and two angles A and B are given: In this case, we use the following formulae to determine the remaining elements \( A + B + C = 180° \Rightarrow C = 180° - A - B \)
\( b = \dfrac{a\sin B}{\sin A} \) and \( c = \dfrac{a\sin C}{\sin A} \Rightarrow \Delta = \dfrac{1}{2}ca\sin B \)

(iv) When two sides a, b and the angle A opposite to one side is given: In this case, we use the following formulae
\( \sin B = \dfrac{b}{a}\sin A \) ..........(i)
\( C = 180° - (A + B),\ c = \dfrac{a\sin C}{\sin A} \)

Special Cases
Case I: When A is an acute angle
(a) If \( a < b\sin A \), there is no triangle. When \( a < b\sin A \), from (i), \( \sin B > 1 \), which is impossible.

From the figure, If AC = b, \( \angle CAX = A \), then perpendicular CN = b sin A. Now taking C as centre, If we draw an arc of radius a then it will never intersect the line AX and hence no triangle ABC can be constructed in this case.
(b) If \( a = b\sin A \), then only one triangle is possible which is right angled at B. When \( a = b\sin A \), then from sine rule, \( \sin B = 1,\ \therefore \angle B = 90° \)
From fig. it is clear that \( CB = a = b\sin A \)
Thus, in this case, only one triangle is possible which is right angled at B.

(c) If \( a > b\sin A \), then three possibilities will arise:
(i) a = b. In this case, from sine rule \( \sin B = \sin A \) ; ∴ B = A or B = 180° − A. But B = 180° − A ⇒ A + B = 180°, which is not possible in a triangle.
∴ In this case, we get \( \angle A = \angle B \).
Hence, if \( b = a > b\sin A \) then only one isosceles triangle ABC is possible in which \( \angle A = \angle B \).

(ii) a > b. In the following figure, Let AC = b, \( \angle CAX = A \), and a > b, also \( a > b\sin A \).
Now taking C as centre, if we draw an arc of radius a, it will intersect AX at one point B and hence only one \( \triangle ABC \) is constructed. Also this arc will intersect XA produced at B′ and \( \triangle AB'C \) is also formed but this Δ is inadmissible (because \( \angle CAB' \) is an obtuse angle in this triangle)
Hence, if a > b and \( a > b\sin A \), then only one triangle is possible.

(iii) b > a (i.e., \( b > a > b\sin A \))
In fig. let AC = b, \( \angle CAX = A \). Now taking C as centre, if we draw an arc of radius a, then it will intersect AX at two points \( B_1 \) and \( B_2 \). Hence if \( b > a > b\sin A \), then there are two triangles.

Case II: When A is an obtuse angle: In this case, there is only one triangle, if a > b

Case III: b > c and B = 90°
Again the circle with A as centre and b as radius will cut the line only in one point. So, only one triangle is possible.

Case IV: b ≤ c and B = 90°
The circle with A as centre and b as radius will not cut the line in any point. So, no triangle is possible.
This is, sometimes called an ambiguous case.

Alternative method: By applying cosine rule, we have \( \cos B = \dfrac{a^2 + c^2 - b^2}{2ac} \)
\( \Rightarrow a^2 - (2c\cos B)a + (c^2 - b^2) = 0 \)
\( \Rightarrow a = c\cos B \pm \sqrt{(c\cos B)^2 - (c^2 - b^2)} \)
\( \Rightarrow a = c\cos B \pm \sqrt{b^2 - (c\sin B)^2} \)
This equation leads to following cases:
Case I: If \( b < c\sin B \), no such triangle is possible.
Case II: Let \( b = c\sin B \), there are further following case.
(a) B is an obtuse angle \( \Rightarrow \cos B \) is negative. There exists no such triangle.
(b) B is an acute angle \( \Rightarrow \cos B \) is positive. There exists only one such triangle.
Case III: Let \( b > c\sin B \). There are further following cases:
(a) B is an acute angle \( \Rightarrow \) cos B is positive.
In this case two values of a will exists if and only if \( c\cos B > \sqrt{b^2 - (c\sin B)^2} \) or c > b. Two such triangle is possible. If c < b, only one such triangle is possible.
(b) B is an obtuse angle \( \Rightarrow \) cos B is negative. In this case triangle will exist if and only if \( \sqrt{b^2 - (c\sin B)^2} > |c\cos B| \Rightarrow b > c \). So, in this case only one such triangle is possible. If b < c there exists no such triangle.

Note: 1. If one side a and two angles B and C are given, then \( A = 180° - (B + C) \) and \( b = \dfrac{a\sin B}{\sin A},\ c = \dfrac{a\sin C}{\sin A} \)
2. If the three angles A, B, C are given. We can only find the ratios of the sides a, b, c by using sine rule (since there are infinite similar triangle possible).

Important Tips

1. A triangle which does not contain a right angle is called an oblique triangle.
2. If triangle is an acute angled triangle means every angle is less than 90°.
3. If triangle is an obtuse angled triangle means one of its angle is greater than 90°.
4. If A and B are complementary angles then A + B = 90°.
5. If A and B are supplementary angle, then A + B = 180°.
6. \( \sum(p - q) = (p - q) + (q - r) + (r - p) = 0,\ \sum p(q - r) = p(q - r) + q(r - p) + r(p - q) = 0 \).
7. \( \sum(p + a)(q - r) = \sum p(q - r) + a\sum(q - r) = 0 \).

Example 47. If in a right angled triangle the hypotenuse is four times as long as the perpendicular drawn to it from opposite vertex, then one of its acute angle is [MP PET 1998]
(A) 15°   (B) 30°   (C) 45°   (D) None of these
Answer:
Ans. (A)
If x is length of perpendicular drawn to it from opposite vertex of a right angled triangle, so, length of hypotenuse \( AB = y_1 + y_2 \) ........(i)
From \( \triangle OCB \), \( y_2 = x\tan\theta \) and from \( \triangle OCA \), \( y_1 = x\cot\theta \)
Put the values in equation (i), then \( AB = x(\tan\theta + \cot\theta) \) .......(ii)
Since, length of hypotenuse = 4 (Length of perpendicular)
∴ \( x(\tan\theta + \cot\theta) = 4x \Rightarrow \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = 4 \)
\( \Rightarrow \sin 2\theta = \dfrac{1}{2} \Rightarrow 2\theta = 30° \Rightarrow \theta = 15° \).
Trick: \( \dfrac{\text{Length of hypotenuse}}{\text{Length of perpendicular drawn from opposite vertex to hypotenuse}} = \dfrac{2}{\sin 2\theta} \)
\( \Rightarrow 4 = \dfrac{2}{\sin 2\theta} \Rightarrow \sin 2\theta = \dfrac{1}{2} \Rightarrow \sin 2\theta = \sin 30° \Rightarrow \theta = 15° \).

Example 48. The number of triangles ABC that can be formed with a = 3, b = 8 and \( \sin A = \dfrac{5}{13} \) is [Roorkee 1998]
(A) 0   (B) 1   (C) 2   (D) 3
Answer:
Ans. (A)
Given, a = 3, b = 8 and \( \sin A = \dfrac{5}{13} \).
∴ \( b\sin A = 8 \times \left(\dfrac{5}{13}\right) = \dfrac{40}{13} > a (= 3) \)
Thus in this case no triangle is possible.

Example 49. In any triangle AB = 2, BC = 4, CA = 3 and D is mid point of BC, then [Roorkee 1995]
(A) \( \cos B = \dfrac{11}{6} \)   (B) \( \cos B = \dfrac{7}{8} \)   (C) AD = 2.4   (D) \( AD^2 = 2.5 \)
Answer:
Ans. (D)
From \( \triangle ABC \), \( \cos B = \dfrac{2^2 + 4^2 - 3^2}{2 \times 2 \times 4} = \dfrac{11}{16} \)
From \( \triangle ABD \), \( \cos B = \dfrac{2^2 + 2^2 - AD^2}{2 \times 2 \times 2} = \dfrac{11}{16} \)
∴ \( \dfrac{11}{16} = \dfrac{2^2 + 2^2 - AD^2}{2 \times 2 \times 2} \)
\( \Rightarrow AD^2 = 2.5 \).

Example 50. If b = 3, c = 4 and \( B = \dfrac{\pi}{3} \), then the number of triangles that can be constructed is [Roorkee 1992]
(A) Infinite   (B) Two   (C) One   (D) Nil
Answer:
Ans. (D)
Here, \( c\sin B = 4\sin\left(\dfrac{\pi}{3}\right) = 2\sqrt{3} > b (= 3) \)
Thus, we have \( b < c\sin B \)
Hence no triangle is possible, i.e., the number of triangles that can be constructed is nil.

Example 51. If two sides of a triangle are \( 2\sqrt{3} \) and \( 2\sqrt{2} \) the angle opposite the shorter side is 45°. The maximum value of the third side is
(A) \( 2 + \sqrt{6} \)   (B) \( \sqrt{2} + \sqrt{6} \)   (C) \( \sqrt{6} - 2 \)   (D) None of these
Answer:
Ans. (B)
Let \( a = 2\sqrt{3},\ b = 2\sqrt{2} \) ∴ B = 45°
∴ \( \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} \Rightarrow \dfrac{2\sqrt{3}}{\sin A} = \dfrac{2\sqrt{2}}{\sin 45°} = \dfrac{c}{\sin C} \) .......(i)
∴ \( \sin A = \dfrac{\sqrt{3}}{2} \Rightarrow A = 60° \) or 120°, so C = 75° or 15°.
From (i) \( c = 4\sin C \), which is largest when C = 75°: \( c = 4\sin(45° + 30°) = \sqrt{2} + \sqrt{6} \) (the other value is \( 4\sin 15° = \sqrt{6} - \sqrt{2} \)).

Example 52. In an ambiguous case, if the remaining angles of the triangles formed with a, b and A be \( B_1, C_1 \) and \( B_2, C_2 \) then \( \dfrac{\sin C_1}{\sin B_1} + \dfrac{\sin C_2}{\sin B_2} = \)
(A) 2 cos A   (B) cos A   (C) 2 sin A   (D) sin A
Answer:
Ans. (A)
In \( \Delta \)'s \( ACB_1 \) and \( ACB_2 \)
\( \dfrac{\sin C_1}{\sin B_1} = \dfrac{AB_1}{AC} = \dfrac{c_1}{b} \) and \( \dfrac{\sin C_2}{\sin B_2} = \dfrac{c_2}{b} \)
∴ \( \dfrac{\sin C_1}{\sin B_1} + \dfrac{\sin C_2}{\sin B_2} = \dfrac{c_1 + c_2}{b} \)
[∵ \( \cos A = \dfrac{b^2 + c^2 - a^2}{2bc} \) or \( c^2 - (2b\cos A)c + (b^2 - a^2) = 0 \) ∴ \( c_1 + c_2 = 2b\cos A \)]
\( \dfrac{\sin C_1}{\sin B_1} + \dfrac{\sin C_2}{\sin B_2} = \dfrac{2b\cos A}{b} = 2\cos A \).