Q: Consider an obtuse angled triangle A B C in which the difference between the largest and the smallest angle is π 2 and whose sides are in arithmetic progression. Suppose that the vertices of this

Let angle A be obtuse angle and let sides be a - d , a , a + d So, plotting the diagram we get, Now let sides be, a − d , a , a + d Also given angle A − C = π 2 And circum radius R = 1 Now by sine rule we get, a + d sin ⁡ A = a sin ⁡ B = a − d sin ⁡ C = 2 ∵ A = π 2 + C ⇒ sin ⁡ A = sin ⁡ π 2 + C ⇒ sin ⁡ A = cos ⁡ C ⇒ a + d 2 = 1 − sin 2 ⁡

Q: Consider an obtuse angled triangle A B C in which the difference between the largest and the smallest angle is π 2 and whose sides are in arithmetic progression. Suppose that the vertices of this

Let angle A be obtuse angle and let sides be a - d , a , a + d So, plotting the diagram we get, Now let sides be, a − d , a , a + d Also given angle A − C = π 2 And circum radius R = 1 Now by sine rule we get, a + d sin ⁡ A = a sin ⁡ B = a − d sin ⁡ C = 2 ∵ A = π 2 + C ⇒ sin ⁡ A = sin ⁡ π 2 + C ⇒ sin ⁡ A = cos ⁡ C ⇒ a + d 2 = 1 − sin 2 ⁡

Q: Consider a triangle P Q R having sides of lengths p , q and r opposite to the angles P , Q and R , respectively. Then which of the following state

(A) Applying cosine rule cos P = q 2 + r 2 − p 2 2 q r = 1 2 q r + r q − p 2 2 q r We know, using A . M . ≥ G . M . q r + r q ≥ 2 ⇒ 1 2 q r + r q ≥ 1 ⇒ 1 2 q r + r q − p 2 2 q r ≥ 1 − p 2 2 q r ⇒ cos P ≥ 1 − p 2 2 q r (B) p + q cos R ≥ q − r cos P + p − r cos Q p cos R + r cos P + q cos R + r cos Q ≥ q cos P + p cos Q using projection formula q + p ≥ r Which is tr

Q: Let x , y and z be positive real numbers. Suppose x , y and z are the lengths of the sides of a triangle opposite to its angles X , Y and Z respectively. If tan X 2 + tan Z

We have, tan X 2 + tan Z 2 = 2 y x + y + z & s = x + y + z = perimeter of a triangle Apply half-angle formula and x + y + z = s then, ⇒ ∆ s s − x + ∆ s s − z = y s ⇒ ∆ = s − x s − z ⇒ ∆ 2 = s s − x s − y s − z = s − x 2 s − z 2 ⇒ y 2 = x 2 + z 2 ⇒ ∠ Y = 90 ° ⇒ ∠ Y = ∠ X + ∠ Z An option B is

Question 1

Consider an obtuse angled triangle A B C in which the difference between the largest and the smallest angle is &#960; 2 and whose sides are in arithmetic progression. Suppose that the vertices of this

Accepted Answer

Let angle  A  be obtuse angle and let sides be  a - d , &#160; a , &#160; a + d So, plotting the diagram we get, Now let sides be,  a &#8722; d , a , a + d Also given angle  A &#8722; C = &#960; 2 And circum radius  R = 1 Now by sine rule we get, a + d sin &#8289; A = a sin &#8289; B = a &#8722; d sin &#8289; C = 2 &#8757; A = &#960; 2 + C &#8658; sin &#8289; A = sin &#8289; &#960; 2 + C &#8658; sin &#8289; A = cos &#8289; C &#8658; a + d 2 = 1 &#8722; sin 2 &#8289;

Question 2

Consider an obtuse angled triangle A B C in which the difference between the largest and the smallest angle is &#960; 2 and whose sides are in arithmetic progression. Suppose that the vertices of this

Accepted Answer

Let angle  A  be obtuse angle and let sides be  a - d , &#160; a , &#160; a + d So, plotting the diagram we get, Now let sides be,  a &#8722; d , a , a + d Also given angle  A &#8722; C = &#960; 2 And circum radius  R = 1 Now by sine rule we get, a + d sin &#8289; A = a sin &#8289; B = a &#8722; d sin &#8289; C = 2 &#8757; A = &#960; 2 + C &#8658; sin &#8289; A = sin &#8289; &#960; 2 + C &#8658; sin &#8289; A = cos &#8289; C &#8658; a + d 2 = 1 &#8722; sin 2 &#8289;

Question 3

Let A B C  be the triangle with A B = 1 , A C = 3  and &#8736; B A C = &#960; 2 . If a circle of radius r &#62; 0  touches the sides A B , A C  and also touches internally the circ

Accepted Answer

Let  A  be the origin  B  on  x - axis,  C  on  y - axis, So on plotting the diagram we get, Now equation of circumcircle is given by  x - 1 2 2 + y - 3 2 2 = 1 2 2 + 3 2 2 = 5 2 &#160; &#160; &#160; &#160; &#160; &#160; &#8943; 1 Required equation of circle touches A B  and A C  and have radius  r  so its equation is given by, x - r 2 + y - r 2 = r 2 &#160; &#160; &#160; &#160; &#160; &#8943; 2 Now given circle in equation 2

Question 4

Consider a triangle  P Q R  having sides of lengths  p , q  and  r  opposite to the angles  P , Q  and  R , respectively. Then which of the following state

Accepted Answer

(A) Applying cosine rule cos P = q 2 + r 2 &#8722; p 2 2 q r = 1 2 q r + r q &#8722; p 2 2 q r We know, using  A . M . &#8805; G . M . q r + r q &#8805; 2 &#8658; &#160; &#160; 1 2 q r + r q &#8805; 1 &#8658; 1 2 q r + r q &#8722; p 2 2 q r &#8805; 1 &#8722; p 2 2 q r &#8658; cos P &#8805; 1 &#8722; p 2 2 q r (B)  p + q cos R &#8805; q &#8722; r cos P + p &#8722; r cos Q p cos R + r cos P + q cos R + r cos Q &#8805; q cos P + p cos Q using projection formula q + p &#8805; r Which is tr

Question 5

In a triangle A B C , let A B = 23 , B C = 3 and C A = 4 . Then the value of  cot A + cot C cot B  is

Accepted Answer

cot A + cot C cot B = cos A sin A + cos C sin C cos B sin B = cos A sin C + cos C sin A sin A sin C &#183; sin B cos B = sin ( A + C ) sin A sin C &#183; sin B cos B = sin ( &#960; - B ) sin A sin C &#183; sin B cos B = sin 2 B sin A sin C cos B Let,  A B = c , &#160; B C = a , &#160; C A = b By using sine rule, we get = b 2 a c cos B = b 2 a c a 2 + c 2 - b 2 2 a c = 2 b 2 a 2 + c 2 - b 2 = 32 9 + 23 - 16 = 32 16 = 2

Question 6

Let x , &#160; y and z be positive real numbers. Suppose x , &#160; y and z are the lengths of the sides of a triangle opposite to its angles X , &#160; Y and Z  respectively. If tan X 2 + tan Z

Accepted Answer

We have,  tan X 2 + tan Z 2 = 2 y x + y + z &#160; &#38; &#160; s = x + y + z = perimeter of a triangle   Apply half-angle formula and  x + y + z = s  then,  &#8658; &#8710; s s &#8722; x + &#8710; s s &#8722; z = y s &#8658; &#8710; = s &#8722; x s &#8722; z &#8658; &#8710; 2 = s s &#8722; x s &#8722; y s &#8722; z = s &#8722; x 2 s &#8722; z 2 &#8658; y 2 = x 2 + z 2 &#8658; &#8736; Y = 90 &#176; &#8658; &#8736; Y = &#8736; X + &#8736; Z An option  B  is

Question 7

In a non-right-angled triangle &#916; P Q R , let p , q , r denote the lengths of the sides opposite to the angles at P , Q , R respectively. The median from R meets the side P Q at S , the perpendicu

Accepted Answer

By sine rule in P Q R p s i n P = q s i n Q = r s i n R = 2 R 3 s i n P = 1 s i n Q = 1 s i n R = 2 1 s i n P = 3 2 and s i n Q = s i n R = 1 2 P = 60 ° or 120 ° , Q = 30 ° and 150 ° ⇒ As P Q R is not a right angle triangle Only possibility is P = 120 ° , Q = R = 30 ° ⇒ P Q R is an isosceles triangle, hence, P E is also a median and ‘ O ’ will be centroid. ( ∆ P Q E and ∆ P R E are congruent and E is mid point of Q R ) ⇒ r = q = 1 , P = 3 Option A : Length of median from R R S = 1 2 2 p 2 + 2 q

Properties of Triangles — Previous Year Questions & Solutions