In a Δ A B C the sides a , b and c are in A . P . Evaluate tan A 2 + tan C 2 tan B 2 .

Using half-angles formula tan A 2 = ( s - b ) ( s - c ) s ( s - a ) tan C 2 = ( s - a ) ( s - b ) s ( s - c ) Adding both of them gives us tan A 2 + tan C 2 = ( s - b ) ( s - c ) s ( s - a ) + ( s - a ) ( s - b ) s ( s - c ) = s - b s s - c s - a + s - a s - c = s - b s 2 s - a - c ( s - a ) ( s - c ) = s - b s b ( s - a ) ( s - c ) = b ( s - b ) s ( s - a ) ( s - b ) ( s - c ) = b ( s - b ) ∆ . . . 1 Now we have to use the formula for inradius to find the relation between semi-perimeter and area r = ( s - b ) tan B 2 = ∆ s ⇒ cot B 2 = s ( s - b ) ∆ = s b × b ( s - b ) ∆ . . . 2 Hence, by 1 and 2 , we have tan A 2 + tan C 2 tan B 2 = b s Now it is given that the sides are in A.P. Hence , 2 b = a + c ⇒ 3 b = a + b + c = 2 s b s = 2 3

The sides of a Δ are in A.P. and its area is 3 5 × (area of an equilateral triangle of the same perimeter). Find the ratio of its sides.

Let the sides be a - d , a , a + d where, d is the common difference. For any triangle, semi-perimeter s has the following relation with sides of the triangle as 2 s = sum of the sides of the triangle = 3 a . ⇒ s = 3 a 2 Area of the triangle is defined as s · ( s - x ) · ( s - y ) · ( s - z ) = 3 a 2 · 3 a 2 - a + d · 3 a 2 - a · 3 a 2 - a - d = 3 a 4 · ( a + 2 d ) · ( a - 2 d ) = 3 a 4 a 2 - 4 d 2 The perimeter of equilateral triangle = Perimeter of the given triangle 3 × one side of equilateral triangle = 3 a ⇒ side of the equilateral triangle = a Area of the equilateral triangle with side a = 3 a 2 4 Now, the ratio of the area of both the triangles is 3 5 . So, a 2 - 4 d 2 a = 3 5 ⇒ 16 a 2 = 100 d 2 a d = 5 2 Ratio of the sides is 5 2 - 1 : 5 2 : 5 2 + 1 = 3 : 5 : 7

In a Δ A B C , R = circumradius and r = inradius. Then the value of a cos A + b cos B + c cos C a + b + c .

Here, we know that if we are able to manipulate the numerator and denominator in terms of the circumradius and inradius, then we can solve the problem. Now, we will treat the numerator and denominator separately Denominator: We know that r = ∆ s ; s = a + b + c 2 ⇒ r = 2 ∆ a + b + c ⇒ a + b + c = 2 ∆ r Now we have to convert the numerator using the formula of circumradius from sine rule a sin A = b sin B = c sin C = 2 R Hence, the numerator becomes R ( 2 sin A cos A + 2 sin B cos B + 2 sin C cos C ) Now we know that sin 2 A = 2 sin A cos A ⇒ a cos A + b cos B + c cos C = R ( sin 2 A + sin 2 B + sin 2 C ) = R ( sin 2 A + 2 sin ( B + C ) cos ( B - C ) ) Now we know that in a triangle A + B + C = π ; B + C = π - A ⇒ sin ( B + C ) = sin ( π - A ) = sinA So, the numerator is R ( 2 sin A cos A + 2 sin A cos ( B - C ) ) = 2 R sin A ( cos A + cos ( B - C ) ) Now applying th property of angles of a triangle, we get cos ( A ) = cos ( π - ( B + C ) ) = - cos ( B + C ) ⇒ 2 R sin A ( cos A + cos ( B - C ) ) = 2 R sin A ( cos ( B - C ) - cos ( B + C ) ) = a ( cos B cos C + sin B sin C - cos B cos C + sin B sin C ) = 2 a sin B sin C So, the numerator is reduced to the form 2 a sin B sin C Now, again applying the circumradius for sin B , we get 2 a b 2 R sin C = 2 R × 1 2 a b sin C = 2 ∆ R ( ∵ ∆ = 1 2 a b sin C ) Now we have numerator = 2 ∆ R and denominator = 2 ∆ r Now dividing both, we get 2 ∆ R 2 ∆ r = r R .

E M B I B E

Skills in Mathematics Trigonometry for JEE Main & Advanced>Chapter 3 - Properties and Solutions of Triangles>Exercise on Level

Amit M Agarwal Solutions for Chapter: Properties and Solutions of Triangles, Exercise 1: Exercise on Level

Author:Amit M Agarwal

Amit M Agarwal Mathematics Solutions for Exercise - Amit M Agarwal Solutions for Chapter: Properties and Solutions of Triangles, Exercise 1: Exercise on Level

Attempt the practice questions on Chapter 3: Properties and Solutions of Triangles, Exercise 1: Exercise on Level with hints and solutions to strengthen your understanding. Skills in Mathematics Trigonometry for JEE Main & Advanced solutions are prepared by Experienced Embibe Experts.

Questions from Amit M Agarwal Solutions for Chapter: Properties and Solutions of Triangles, Exercise 1: Exercise on Level with Hints & Solutions

MEDIUM

JEE Main

IMPORTANT

Skills in Mathematics Trigonometry for JEE Main & Advanced>Chapter 3 - Properties and Solutions of Triangles>Exercise on Level>Q 23

If in a $Δ A B C,$ the incentre is the middle point of the median $A D$ . Then measure $\cos A$ .

HARD

JEE Main

IMPORTANT

Skills in Mathematics Trigonometry for JEE Main & Advanced>Chapter 3 - Properties and Solutions of Triangles>Exercise on Level>Q 24

In a $Δ A B C$ the sides $E$ and $E \equiv (\frac{x_{1} + \frac{x_{2} + x_{3}}{2}}{2}, \frac{y_{1} + \frac{y_{2} + y_{3}}{2}}{2})$ are in $A . P .$ Evaluate $(\tan \frac{A}{2} + \tan \frac{C}{2}) \tan \frac{B}{2}$ .

HARD

JEE Main

IMPORTANT

Skills in Mathematics Trigonometry for JEE Main & Advanced>Chapter 3 - Properties and Solutions of Triangles>Exercise on Level>Q 25

The sides of a $Δ$ are in $A.P.$ and its area is $\frac{3}{5} \times$ (area of an equilateral triangle of the same perimeter). Find the ratio of its sides.

HARD

JEE Main

IMPORTANT

Skills in Mathematics Trigonometry for JEE Main & Advanced>Chapter 3 - Properties and Solutions of Triangles>Exercise on Level>Q 26

If $A D, B E$ and $C F$ are the medians of a $Δ A B C,$ then evaluate $(A D^{2} + B E^{2} + C F^{2}) : (B C^{2} + C A^{2} + A B^{2})$ .

MEDIUM

JEE Main

IMPORTANT

Skills in Mathematics Trigonometry for JEE Main & Advanced>Chapter 3 - Properties and Solutions of Triangles>Exercise on Level>Q 27

$A D$ is a median of the $Δ A B C$ . If $A E$ and $A F$ are medians of the triangles $A B D$ and $A D C$ , respectively, and $A D = m_{1}, A E = m_{2}, A F = m_{3},$ then $\frac{a^{2}}{8} = p {m_{2}}^{2} + q {m_{3}}^{2} - r {m_{1}}^{2}$ . Find $(p + q) r$ .