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Tricks & Tips Of Properties of Triangles For NDA

Properties of triangle

Q 2753880744

Consider the following for triangle ABC:

1. `sin((B+C)/2) = cos (A/2)`

2. `tan (( B+C)/2) = cot(A/2)`

3. `sin (B+C) = cos A`

4. `tan (B+C) = - cot A`

Which of the above are correct?


NDA Paper 1 2017
(A)

1 and 3

(B)

1 and 2

(C)

1 and 4

(D)

2 and 3

Solution:

For `Delta ABC`

`A+ B+ C= pi`

1. `sin ((B+C)/2)= sin (pi/2 - A/2) = cos((A/2))`

2. `tan ((B+C)/2) = tan (pi/2 - A/2) = cot (A/2)`

3. `sin (B+C) = sin (pi-A) = sin A`

4. `tan (B+C) = tan (pi- A) =-tan A`
Correct Answer is `=>` (B) 1 and 2
Q 2410701610

`ABC` is a triangle right angled at `B`. The hypotenuse (`A C`)
is four times the perpendicular (`BD`) drawn to it from the
opposite vertex and `AD < DC`.

What is one of the acute angle of the triangle?
NDA Paper 1 2011
(A)

`15^o`

(B)

`30^o`

(C)

`45^o`

(D)

None of these

Solution:

Let `BD = p` and `DE = x`

`=> AC=4p`

Let `E` be a mid-point of line `AC`.

Then,` AE = EC =BE =2p`

In `Delta BDE,`

`(BE)^2=(BD)^2+(ED)^2`

`(2p)^2=(p)^2+(x)^2`

`4p^2=p^2+x^2`

`x^2=3p^2`

`x=sqrt 3p`

Now, `AD= 2p- x = 2p- p sqrt 3`

`DC =2p+ x=2p+ p sqrt 3`

In `Delta BAD`

`tan A=(BD)/(AD)=p/(2P-P sqrt 3)=1/(2-sqrt 3) xx (2+sqrt 3)/(2+sqrt 3)`

`=(2+sqrt 3)/1=tan 75^o`

`=> A=75^o`

`tan alpha=(AD)/(BD)=(2p-sqrt 3 p)/p=2-sqrt 3=tan 15^o`

`=> alpha=15^o`

In `Delta BDC`

`tan C=(BD)/(CD)=p(2p+p sqrt 3)=(2-sqrt 3)/(4-3)`

`=> tan C=2-sqrt 3=tan 15^o=> C=15^o`

`tan beta=(CD)/(BD)=(2p+sqrt 3 p)/p=2+sqrt 3=tan 75^o`

`=> beta=75^o`

Acute angle ` =15^o`
Correct Answer is `=>` (A) `15^o`
Q 2420491311

If one of the angles of a triangle is `1/2` radian and
the other is `99^0`, then what is the third angle in
radian measure?
NDA Paper 1 2010
(A)

`(9pi-10)/pi`

(B)

`(90pi-100)/(7pi)`

(C)

`(90pi-10)/pi`

(D)

None of these

Solution:

Let `angleA = 1/2 rad , angleB = 99^0 = (99^0xxpi)/(180^0) = (11pi)/(20)`


We know that, `angleA+angleB+angleC = pi`


`=> 1/2+(11pi)/20+angleC = pi`


`therefore angleC = pi-(11pi)/(20)-1/2 = (9pi-10)/(20)`
Correct Answer is `=>` (D) None of these
Q 2311780620

Consider the following statements
1. If `ABC` is an equilateral triangle, then `3 tan (A +B) tan C = 1.`

2. If `ABC` is a triangle in which `A= 78°, B = 66°,` then `tan(A/2 + C) < tan A`

3. If `ABC` is any triangle , then `tan ((A+B)/2) sin (C/2) < cos (C/2)`

Which of the above statements is/are correct?
NDA Paper 1 2016
(A)

Only 1

(B)

Only 2

(C)

1 and 2

(D)

2 and 3

Solution:

1. Since, `ABC` is an equilateral triangle.
`:. angle A = angleB = angleC = 60^o`

Now, consider
`3 tan (A+ B) tan C = 3 tan 120° tan 60°`

`= 3 tan (180° - 60°) tan 60°`

`=(-3) tan 60^o tan 60^o = (-3 ) * sqrt 3 * sqrt3 = -9`

2. Consider , `tan (A/2 +C) = tan (78^o/2 +36^o)`

`[because angleC = 180^o - angleA - angleB = 180^o - 78^o - 66^o = 36^o]`

`= tan 75^0 < tan 78^o`

[ `because` value of tan e increases, as e varying from 0° to 90°]
`=tan A`

Hence , `tan (A/2 + C) < tan A`
Correct Answer is `=>` (B) Only 2
Q 2309891718

If the angles of a triangle are in AP and the least
angle is `30°`, then what is the greatest angle (in
radian)?
NDA Paper 1 2012
(A)

`pi/2`

(B)

`pi/3`

(C)

`pi/4`

(D)

`pi`

Solution:

Let the angles of a triangle are `a, a+ d` and `a+ 2d`.

(since, given that angles of a triangle are in A..P)

Given that, `a = 30°`

`·: a + a + d + a + 2d = 180°`

(since, the sum of internal angles of a triangle is `180^o`)

`=> 3a+3d=180°`

`=> 3 xx 30° + 3d = 180°`

`=>90° + 3d = 180°`

`=>3d= 90^o`

`=>d = 30°`

So, the angles of triangle are `30°, 60°` and `90^o`.

Hence, greatest angle `= 90°= pi/2`
Correct Answer is `=>` (A) `pi/2`
Q 2470523416

If median of the `triangle ABC` through `A` is perrendicular
to `BC`, then whtch one of the following is correct?
NDA Paper 1 2007
(A)

`tan A + tan B = 0`

(B)

`tan B - tan C = 0`

(C)

`tan C + 2 tan A = 0`

(D)

`tan B + tan C = 0`

Solution:

Let `BC=a`

`:. BD=CD=a/2`

In `triangle ABD, tan B=(AD)/(BD)=(AD)/(a//2)`

`=> tan B=(2 AD)/a`...............(i)

In `triangle ACD, tan C=(AD)/(CD)=(AD)/(a//2)`

`=> tan C=(2 AD)/a`.............(ii)

From Eqs. (i) and (ii),

`tan B=tan C`

`=> tan B-tan C=0`
Correct Answer is `=>` (B) `tan B - tan C = 0`
Q 2166580475

Consider a `Delta ABC` in which `cos A+ cos B + cos C = sqrt(3) sin quad (pi)/(3)`

What is the value of `sin (A/2) sin (B/2) sin (C/2) ?`
NDA Paper 1 2016
(A)

`1/2 `

(B)

`1/4 `

(C)

`1/8 `

(D)

`1/16 `

Solution:

Given, in `DeltaABC,`

`cos A + cos B + cos C = sqrt(3) sin quad (pi)/3`

`=> cos A + cos B + cos C = sqrt(3) xx sqrt(3)/2 = 3/2`

We know that,

`cos A + cos B + cos C = 1+4 sin (A/2) - sin (B/2) sin (C/2)`

` => 1 + 4 sin (A/2) sin (B/2) sin (C/2) =3/2`

`=> 4 sin (A/2) sin (B/2) sin (C/2) =3/2 -1 =1/2 `

` => sin (A/2) sin (B/2) sin (C/2) =1/8 `
Correct Answer is `=>` (C) `1/8 `
Q 2460734615

If `tan^(-1) 2` and `tan^(-1) 3` are two angles of a triangle,
then what is the third angle?
NDA Paper 1 2012
(A)

`tan^(-1) 2`

(B)

`tan^(-1) 4`

(C)

`pi/4`

(D)

`pi/3`

Solution:

Given that, `tan^(-1)2` and `tan^(-1) 3` are two angles of a
triangle. Let `alpha` be the third angle of the triangle, then -
In `Delta ABC, tan^(-1) 2 + tan^(-1) 3 + alpha= 180°`

`=> tan^(-1) ((2+3)/(1-2 *3))+alpha=180^o`

`[ ∵ tan^(-1) x+tan^(-1)y=tan^(-1) ((x+y)/(1-xy))]`

`=>tan^(-1) (5/(1-6))+alpha=180^o=> tan^(-1) (5/(-5))+alpha=180^o`

`=> tan^(-1) (-1)+alpha=180^o=>alpha=180^o-tan^(-1)(-1)=pi-(3 pi)/4`

`:. alpha=pi/4`
Correct Answer is `=>` (C) `pi/4`
Q 2335278162

If in a `Delta ABC, cos B = (sin C)/(2 sin A)`, then the triangle is
BITSAT Mock
(A)

right-angled

(B)

equilateral

(C)

isosceles

(D)

right-angled isosceles

Solution:

`sin C = 2 sin A cos B`

`= sin (A + B) + sin (A - B)`

`= sin C + sin (A - B)`

`=> sin (A - B) = 0 => A = B`

i.e., the triangle is isosceles.
Correct Answer is `=>` (C) isosceles
Q 2268467305

If in a `DeltaABC, a^2 + b^2 + c^2 = ca + ab sqrt(3)`,

then the triangle
BITSAT Mock
(A)

is equilateral

(B)

is right-angled isosceles

(C)

has angles `30^(circ), 60^(circ)` and `90^(circ) `

(D)

None of these

Solution:

`a^2 + b^2 + c^2 - ca - ab sqrt(3) = 0`

Regrouping,

`(b- (a sqrt(3))/2)^2 + (c- a/2)^2 =0`

`:. a : b : c = 2 : sqrt(3) : 1`.

Hence the triangle has angles

`30^(circ), 60^(circ)` and `90^(circ)`.
Correct Answer is `=>` (C) has angles `30^(circ), 60^(circ)` and `90^(circ) `
Q 2420701611

`ABC` is a triangle right angled at `B`. The hypotenuse (`A C`)
is four times the perpendicular (`BD`) drawn to it from the
opposite vertex and `AD < DC`.

What is `angle ABD` ?
NDA Paper 1 2011
(A)

`15^o`

(B)

`30^o`

(C)

`45^o`

(D)

None of these

Solution:

Let `BD = p` and `DE = x`

`=> AC=4p`

Let `E` be a mid-point of line `AC`.

Then,` AE = EC =BE =2p`

In `Delta BDE,`

`(BE)^2=(BD)^2+(ED)^2`

`(2p)^2=(p)^2+(x)^2`

`4p^2=p^2+x^2`

`x^2=3p^2`

`x=sqrt 3p`

Now, `AD= 2p- x = 2p- p sqrt 3`

`DC =2p+ x=2p+ p sqrt 3`

In `Delta BAD`

`tan A=(BD)/(AD)=p/(2P-P sqrt 3)=1/(2-sqrt 3) xx (2+sqrt 3)/(2+sqrt 3)`

`=(2+sqrt 3)/1=tan 75^o`

`=> A=75^o`

`tan alpha=(AD)/(BD)=(2p-sqrt 3 p)/p=2-sqrt 3=tan 15^o`

`=> alpha=15^o`

In `Delta BDC`

`tan C=(BD)/(CD)=p(2p+p sqrt 3)=(2-sqrt 3)/(4-3)`

`=> tan C=2-sqrt 3=tan 15^o=> C=15^o`

`tan beta=(CD)/(BD)=(2p+sqrt 3 p)/p=2+sqrt 3=tan 75^o`

`=> beta=75^o`

`:. angle ABD = 15°`
Correct Answer is `=>` (A) `15^o`
Q 2420001811

`ABC` is a triangle right angled at `B`. The hypotenuse (`A C`)
is four times the perpendicular (`BD`) drawn to it from the
opposite vertex and `AD < DC`.

What is `AD: DC ` equal to?

NDA Paper 1 2011
(A)

`(7-2 sqrt 3) :1`

(B)

`(7 -4 sqrt 3) :1`

(C)

`1:2`

(D)

None of these

Solution:

Let `BD = p` and `DE = x`

`=> AC=4p`

Let `E` be a mid-point of line `AC`.

Then,` AE = EC =BE =2p`

In `Delta BDE,`

`(BE)^2=(BD)^2+(ED)^2`

`(2p)^2=(p)^2+(x)^2`

`4p^2=p^2+x^2`

`x^2=3p^2`

`x=sqrt 3p`

Now, `AD= 2p- x = 2p- p sqrt 3`

`DC =2p+ x=2p+ p sqrt 3`

In `Delta BAD`

`tan A=(BD)/(AD)=p/(2P-P sqrt 3)=1/(2-sqrt 3) xx (2+sqrt 3)/(2+sqrt 3)`

`=(2+sqrt 3)/1=tan 75^o`

`=> A=75^o`

`tan alpha=(AD)/(BD)=(2p-sqrt 3 p)/p=2-sqrt 3=tan 15^o`

`=> alpha=15^o`

In `Delta BDC`

`tan C=(BD)/(CD)=p(2p+p sqrt 3)=(2-sqrt 3)/(4-3)`

`=> tan C=2-sqrt 3=tan 15^o=> C=15^o`

`tan beta=(CD)/(BD)=(2p+sqrt 3 p)/p=2+sqrt 3=tan 75^o`

`=> beta=75^o`

`AD : DC= (7- 4sqrt 3) : 1`
Correct Answer is `=>` (B) `(7 -4 sqrt 3) :1`
Q 2430612512

If in a `triangle ABC, cos B =(sin A) // (2 sin C)`, then the
triangle is
NDA Paper 1 2009
(A)

isosceles triangle

(B)

equilateral triangle

(C)

right angled triangle

(D)

scalene triangle

Solution:

Using sine rule

`(sin A)/a=(sin B)/b=(sin C)/c=k`

Given, `cos B=(sin A)/(2 sin C)`

`=> (a^2+c^2-b^2)/(2ac)=a/(2c)`

`=>a^2+c^2-b^2=a^2`

`=>b^2=c^2`

`:. b=c`

Hence, it is an isosceles triangle.
Correct Answer is `=>` (A) isosceles triangle
Q 2126680571

Consider a `Delta ABC` in which `cos A+ cos B + cos C = sqrt(3) sin quad (pi)/(3)`

What is the value of
`cos ((A+B)/2) cos ((B+C)/2) cos ((C+A)/2) ?`
NDA Paper 1 2016
(A)

`1/4`

(B)

`1/2`

(C)

`1/16`

(D)

None of these

Solution:

`cos ((A+B)/2) cos ((B+C)/2) cos ((C+A)/2) `

` = cos (90^(0) - C/2) cos ( 90^(0)- A/2) cos (90^(0)- B/2) `

` [ ∵ A + B + C = 180^0]`

` sin (C/2) sin (A/2) sin (B/2) `

` sin (A/2) sin (B/2) sin (C/2) `

` = 1/8` [from Q. No. 51]
Correct Answer is `=>` (D) None of these
Q 2241501423

If `a, b` and `c` are the sides of a `Delta ABC`, then `a^(1/p) + b^(1/p) - c^(1/ p)`,
where `p > 1`, is
NDA Paper 1 2015
(A)

always negative

(B)

always positive

(C)

always zero

(D)

positive, if `1 < p < 2` and negative, if `p > 2`

Solution:

Since, `a, b, c` are positive numbers.

`=> a^(1/p) , b^(1/p) , c^(1/ p)` are positive numbers.

` => a^(1/p) + b^(1/p) > c^(1/ p)`

[ `∵` sum of any two sides must be greater than third side]

` => a^(1/p) + b^(1/p) - c^(1/ p) > 0`
Correct Answer is `=>` (B) always positive
Q 1780256117

In a `Delta ABC`, if `sin A -cos B = cos C`, then what is `B` equal to?
NDA Paper 1 2014
(A)

`pi`

(B)

`pi/3`

(C)

`pi/2`

(D)

`pi/4`

Solution:

In a `Delta ABC`, we have

`=> sin A - cos B = cos C => sin A = cos B + cos C`

`=> 2 sin (A/2) . cos .A/2 =2 cos ((B +C)/2) . cos ((B - C)/2)`

`[∵ sin2A = 2sin A . cos A`

and `cos B +cos C = 2 cos ((B +C)/2) . cos ((B - C)/2) ]`

`=> 2 sin (A/2) . cos (A/2) =2 cos ( 90^0 - A/2 ) . cos ((B - C)/2) `

` [ ∵ A+B+C=180^0 => ((B - C)/2) = 90^0 - A/2 ]`

`=> 2 sin (A/2) . cos (A/2) = 2. sin (A/2) . cos ((B - C)/2) `

` => cos (A/2) = cos ((B - C)/2) `

`=> A/2 = (B-C)/2 `

`=> A + C = B ` .........(1)

also `A + C =180^0 - B` .........(2)

` 180^0 - B = B`

` => 2B = 180^0`

`:. B = 90^0`
Correct Answer is `=>` (C) `pi/2`

Sine and cosine Rule

Q 2410712619

In a `triangle ABC, a+b=3(1+sqrt3) cm` and `a-b=3(1-sqrt 3) cm` . If angle `A` is `30^o` , then what is the angle `B` ?
NDA Paper 1 2009
(A)

`120^o`

(B)

`90^o`

(C)

`75^o`

(D)

`60^o`

Solution:

Given, `a + b = 3(1 + sqrt 3)` ...................(i)

and `a - b = 3(1 - sqrt 3)`...................(ii)

On solving, we get

`a/(sin A)=b/(sin B)`

`=> 3/(sin 30^o)=(3 sqrt3)/(sin B)`

`=> sin B=sqrt 3 xx 1/2 =sin 60^o`

`:. B=60^o`
Correct Answer is `=>` (D) `60^o`
Q 2460301215

If the sides of a triangle are in the ratio
`2 : sqrt 6 :1 + sqrt 3` , then what is the smallest angle of
the triangle?
NDA Paper 1 2011
(A)

`75^o`

(B)

`60^o`

(C)

`45^o`

(D)

`30^o`

Solution:

Let `a, b` and `c` be the sides of `Delta ABC`, respectively.

Given, `a : b : c=2 : sqrt 6 : (1+ sqrt 3)`...............(i)

We know that, by sine law

`a/(sin A)=b/(sin B)=c/(sin C)=k` (say)

From Eq (i),

`k sin A : k sin B : ksin C=2: sqrt 6: (1+sqrt 3)`

`=> sin A : sin B : sin C =sqrt (2/3) :1 :(1+sqrt 3)/(sqrt 6)`

`=> 1/sqrt 2: sqrt 3/2 :(1+sqrt 3)/(2 sqrt 2)`

`=sin 45^o : sin 60^o : sin 75^o`

`=> A: B: C=45^o :60^o :75^o`

So, the required smallest angle of `Delta ABC` is `45°`.
Correct Answer is `=>` (C) `45^o`
Q 2389191917

In any `DeltaABC`, the sides are `6 cm, 1 0 cm` and
`14 cm`. Then, the triangle is obtuse angled with
the obtuse angle equal to
NDA Paper 1 2011
(A)

`150^o`

(B)

`135^o`

(C)

`120^o`

(D)

`105^o`

Solution:

Given in `Delta ABC, a= 6cm, b = 10cm, c = 14cm`

Since, `C` is the long side among three sides and the obtuse angle
of `Delta ABC` is the corresponding angle of side `c`.

By cosine law,

`cos C=(a^2+b^2-c^2)/(2ab)=(6^2+10^2-14^2)/(2 * 6 * 10)`

`=> cos C=(36+100-196)/(2 * 6 * 10)=(136-196)/(2 * 6 *10)`

`=> cos C=-1/2=cos (2 pi)/3`

`:. C=(2 pi)/3=120^o`
Correct Answer is `=>` (C) `120^o`
Q 1658178904

In a `Delta ABC, a =(1 + sqrt(3))cm, b =2 cm` and `angleC = 60^0`,
then the other two angles are
NDA Paper 1 2015
(A)

`45^0` and `75^0`

(B)

`30^0` and `90^0`

(C)

`105^0` and `15^0`

(D)

`100^0` and `20^0`

Solution:

Here, `a =(1 + sqrt(3)) , b =2 , angleC = 60^0`

`:. cos C = (a^2 +b^2 - c^2)/(2ab)`

` => cos 60^0 = ( ( 1+ sqrt(3) )^2 + (2)^2 -c^2)/(2(1 + 3)^2)`

` => 1/2 xx 2 xx 2 xx(1 + sqrt(3))= 1 + 3+ 2 sqrt(3) + 4 - c^2`

`=> 2 + 2 sqrt(3) = 4 + 2 sqrt(3) + 4 - C^2`

` => c^ 2 = 8 + 2sqrt(3) - 2 - 2 sqrt(3) = 6`

` => c = sqrt(16)`

`:. cos A = (a^2 +b^2 - c^2)/(2ab)`

` = ( 4 + 6 - ( 1 + sqrt(3) )^2)/(2 xx 2 xx sqrt(6))`

` = (10 - 1 - 3 - 2 sqrt(3))/(4 sqrt(6)) = ( 6 - 2 sqrt(3))/(4 sqrt(6))`

` = ( 3 - sqrt(3) )/(2 sqrt(6)) = ( 3 sqrt(6) - sqrt(18))/(12)`

` = (sqrt (3) -1 ) / ( 2 sqrt(2)) = cos 75 ^0`

`:. A = 75 ^0 , B = 45 ^0 ,C = 60 ^0`
Correct Answer is `=>` (A) `45^0` and `75^0`
Q 2480212117

In a `triangleABC, BC = sqrt(39), AC = 5` and `AB = 7`. What
is the measure of the angle `A'`?
NDA Paper 1 2010
(A)

`pi/4`

(B)

`pi/3`

(C)

`pi/2`

(D)

`pi/6`

Solution:

Here, `a = sqrt(39), b = 5` and `c = 7`

By cosine rule, `cos A =( b^2 + c^2- a^2)/(2bc)`

`=(25 + 49- 39)/(2 xx 5 xx 7)`

`= 1/2 =cos \ pi/3`

`=> A= pi/3`
Correct Answer is `=>` (B) `pi/3`
Q 2410423310

In a `triangle ABC, b = sqrt 3 cm, c =1 cm` and `angle A =30^o`
what is the value of `a` ?
NDA Paper 1 2008
(A)

`sqrt 2` cm

(B)

`2` cm

(C)

`1` cm

(D)

`1/2` cm

Solution:

Here, `b = sqrt 3cm, c = 1 cm` and `angle A = 30°`

We know that,

`cos A=(b^2+c^2-a^2)/(2bc)`

`=>cos 30=((sqrt 3)^2+(1)^2-a^2)/(2 sqrt 3 *1)`

`=> sqrt 3/2=(3+1-a^2)/(2 sqrt 3)`

`=>3=4-3=1`

`:. a=1cm`
Correct Answer is `=>` (C) `1` cm
Q 2539080812

In `Delta ABC`, if ` a/(b^2 - c^2) + c/(b^2 - a^2) = 0`, then `B` is equal to


BCECE Mains 2015
(A)

`pi/2`

(B)

`pi//4`

(C)

`2pi//3`

(D)

`pi//3`

Solution:

We have,

`a/(sin A) = b/(sin B) = c/(sin C) = k`

`:. a/(b^2 - c^2) + c/(b^2 - a^2) = 0`

`=> (k sin A)/(k^2 (sin^2 B - B sin^2 C)`

`+ (k sin C)/(k^2 (sin^2 B - B sin^2 A) = 0`

` => (sin A)/(sin (B + C) sin (B - C))`

` + ( sin C)/(sin (B + A) sin (B - A)) = 0`

` => 1/(sin (B -C)) + 1/(sin (B -A)) = 0`

` => sin (B - A) + sin (B -C) = 0`

`=> sin (A - B) = sin (B -C)`

`=> A - B = B - C`

`=> A + C= 2B => B = 60^0`
Q 2542212133

Let p, q and r be the sides opposite to the angles P, Q and R, respectively in a `Delta PQR`. If `r^2 sin P sin Q = pq`, then the triangle is
WBJEE 2012
(A)

equilateral

(B)

acute angled but not equilateral

(C)

obtuse angled

(D)

right angled

Solution:

We know that in `Delta ABC`

`a/(sin A) = b/(sin B) = c/(sin C) = 2R`

`:. r^2 sin P sin Q = pq`

`=> r^2 · p/(2R_1) · q/(2 R_1) = pq`, where `R_1` is circumradius

of `Delta PQR`.

`=> r^2 = 4R_1^2`

`=> r = 2R_1`

` => 2R_1 sin R = 2R_1`

`=> R_1 = 90°`

`:. Delta PQR` is right angled.
Correct Answer is `=>` (D) right angled
Q 2476267176

If in `DeltaABC, cos A+ cosB + cosC = 3/2`, then
triangle `Delta` is
UPSEE 2013
(A)

right angled

(B)

isosceles

(C)

acute

(D)

equilateral

Solution:

`:. cos A+ cos B + cos C = 3/2`

`=> (b^2 +c^2 -a^2)/(2bc) + (a^2 + c^2 -b^2)/(2ac)+ (a^2 + b^2 -c^2)/(2ab) =3/2`

`=> (a+b -c) (a-b)^2 + (b+c -a) (b-c)^2`

`+ (c+a-b) (c-a)^2 =0`

As we know, `a+ b > c, b + c >a, c + a > b`


`=> a = b = c`

Hence, triangle is an equilateral.
Correct Answer is `=>` (D) equilateral
Q 1732834732

Consider the following statements

I. There exists no `Delta ABC` for which `sin A + sin B = sin C`.

II. If the angles of a triangle are in the ratio `1 : 2 : 3`, then
its sides will be in the ratio `1 : sqrt(3) : 2`.

Which of the above statements is/are correct?
NDA Paper 1 2014
(A)

Only 1

(B)

Only 2

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

I. Given that, `sin A+ sin B = sin C`

` (ak) + (bk) = (ck)`

`=> a+ b = c`

` (∵` by sine law , ` (sin A)/a = (sin B)/b = ( sin C)/b = k)`

i.e., the sum of two sides of `Delta ABC` is equal to the third side

but it is not possible, because by triangle inequality 'the

sum of the length of two sides of a triangle is always

greater than the length of the third side'.

Hence, there exists no `Delta ABC` for which `sin A+ sin B =sin C`.

IL Given that.

the ratio of the angles of a triangle are `A : B : C = 1 : 2 : 3`.

Let `A = alpha , B = 2 alpha`. and `C = 3alpha`

We know that,

`A+ B + C = 180^0`

(since, sum of all interior angles of a triangle is `180^0`)

`=> alpha + 2alpha + 3alpha = 180^0`

` => 6 alpha = 180^0 => alpha = 30^0`

`:. A = 30^0 , B = 60^0` and `C = 90^0`

So, the required ratio in its sides,

` a : b : c = sin A :sin B :sin C` (by sine rule)

` = sin 30^0 :sin 60^0 :sin 90^0`

` = 1/2 : sqrt(3)/2 : 1 = 1: sqrt(3) : 2`
Correct Answer is `=>` (C) Both 1 and 2
Q 1730045812

In a ` Delta ABC`, if `c =2, A =45^0` and `a = 2 sqrt(2)`, then wll.at is `C`
equal to?
NDA Paper 1 2014
(A)

`30^0`

(B)

`15^0`

(C)

`45^0`

(D)

None of these

Solution:

By sine rule,

` a/(sin A) = b/( sin B) = c/( sin C)`

`:. a/(sin A) = c/( sin C)`

` => sin C = ( c . sin A) /a = ( 2 . sin 45^0)/(2 sqrt(2))`

` = 1/sqrt(2) . 1/sqrt(2) = 1/2 = sin 30^0`

`:. C = 30^0`
Correct Answer is `=>` (A) `30^0`
Q 2389191917

In any `DeltaABC`, the sides are `6 cm, 1 0 cm` and
`14 cm`. Then, the triangle is obtuse angled with
the obtuse angle equal to
NDA Paper 1 2011
(A)

`150^o`

(B)

`135^o`

(C)

`120^o`

(D)

`105^o`

Solution:

Given in `Delta ABC, a= 6cm, b = 10cm, c = 14cm`

Since, `C` is the long side among three sides and the obtuse angle
of `Delta ABC` is the corresponding angle of side `c`.

By cosine law,

`cos C=(a^2+b^2-c^2)/(2ab)=(6^2+10^2-14^2)/(2 * 6 * 10)`

`=> cos C=(36+100-196)/(2 * 6 * 10)=(136-196)/(2 * 6 *10)`

`=> cos C=-1/2=cos (2 pi)/3`

`:. C=(2 pi)/3=120^o`
Correct Answer is `=>` (C) `120^o`

Trigonometric ratios of half Angels, Tangent formulae and projection formulae

Q 2271412326

Consider a `Delta ABC` satisfying
`2a sin^2 (C/2) + 2c sin^2 ( A/2) = 2a + 2c - 3b`
The sides of the triangle are in
NDA Paper 1 2015
(A)

`GP`

(B)

`AP`

(C)

`HP`

(D)

Neither in `GP` nor `AP` nor in `HP`

Solution:

We have, `2a sin^2 (C/2) + 2c sin^2 ( A/2) = 2a + 2c - 3b`

`=> a(1 - cos C) + c(1 - cos A)= 2a + 2c- 3b`

`=> (a+ c) - (a cos C + c cos A) = 2a+ 2c-3b`

`=> a + c- b = 2a + 2c- 3b`

`[ ∵ b =a cos C + c cos A]`

`=> - a - c =- 2b => a + c = 2b `

Hence, `a, b, c` are in `AP`.
Correct Answer is `=>` (B) `AP`
Q 2400223118

If `ABC` is a triangle in which `AB = 6 cm, BC = 8
cm` and `CA = 10 cm`, then what is the value of
`cot (A // 4)` ?
NDA Paper 1 2008
(A)

`sqrt 5 -2`

(B)

`sqrt 5+2`

(C)

`sqrt 3-1`

(D)

`sqrt 3+1`

Solution:

Here, `a= 8, b = 10` and `c = 6`

Now, `S=(a+b+c)/2`

`=(8+10+6)/2=12`

`:. tan A/2=sqrt(((12-10)(12-6))/(12(12-8)))`

`=sqrt (1/4)=1/2`

`[:. tan (A/2) =sqrt(((s-b)(s-c))/(s(s-a))) text(and) s=1/2 (a+b+c)]`

`=> cot A/2=2`

Now, `cot(A/4+A/4)=(cot^2 A/4-1)/(2 cot A/4)`

`[:. cot (A+B)=(cot A * cot B-1)/(cot A+cot B)]`

`cot(A/2)=(cot^2A/4-1)/(2 cot A/4)`

`:. 2=(x^2-1)/(2x)`

`=> x^2-4x-1=0`

`:. x=(4 pm sqrt(16+4))/2`

`=(4 pm 2 sqrt 5)/2=2 pm sqrt 5`

So, `cot(A/4)=2 + sqrt 5` or `2-sqrt 5`
Correct Answer is `=>` (B) `sqrt 5+2`
Q 2271412326

Consider a `Delta ABC` satisfying
`2a sin^2 (C/2) + 2c sin^2 ( A/2) = 2a + 2c - 3b`
The sides of the triangle are in
NDA Paper 1 2015
(A)

`GP`

(B)

`AP`

(C)

`HP`

(D)

Neither in `GP` nor `AP` nor in `HP`

Solution:

We have, `2a sin^2 (C/2) + 2c sin^2 ( A/2) = 2a + 2c - 3b`

`=> a(1 - cos C) + c(1 - cos A)= 2a + 2c- 3b`

`=> (a+ c) - (a cos C + c cos A) = 2a+ 2c-3b`

`=> a + c- b = 2a + 2c- 3b`

`[ ∵ b =a cos C + c cos A]`

`=> - a - c =- 2b => a + c = 2b `

Hence, `a, b, c` are in `AP`.
Correct Answer is `=>` (B) `AP`
Q 2573601546

If in a `Delta ABC, sin A, sin B, sin C` are in AP,
then
WBJEE 2011
(A)

the altitudes are in AP

(B)

the altitudes are in HP

(C)

the angles are in AP

(D)

the angles are in HP

Solution:

If `p_1, p_2` and `p_3` are the altitudes of a triangle.

Then, `Delta 1/2 ap_1 =1/2 bp_2 = 1/2 c p_3`

`=> a= (2 Delta)/p_1 , b =(2 Delta)/p_2, c= (2 Delta)/p_3`

Since, `sin A, sin B` and `sin C` are in AP.

`:. sin B = (sin A + sin C)/2`

`=> b= (a+c)/2`

`=> (2 Delta)/(p_2) = ( ( (2 Delta)/(p_1) ) + ((2 Delta)/(p_3)) )/2`

`=> 2/p_2 = (p_3 +p_1)/(p_1 p_3)`

`=> p_2 = (2p_1 p_3)/(p_1 + p_3)`

Hence, altitudes are in AP.
Correct Answer is `=>` (A) the altitudes are in AP
Q 2448178003

If in a `Delta ABC`, the altitudes from the
vertices `A, B, C` on opposite sides are in HP
then `sin A, sin B, sin C` are in
BCECE Stage 1 2016
(A)

HP

(B)

Arithmetico-Geometric Progression

(C)

AP

(D)

GP

Solution:

In `Delta BAD`,

`cos (90-B)=(AD)/c`

`AD=c sin B`

Similarly, `BE= a sin C` and `CF = bsin A`

Since, `AD, BE, CF` are in HP

`:. csin B, a sin C, bsin A` are in HP

`1/(sin C sin B), 1/(sin A sin C), 1/(sin B sin A)` are in AP

`sin A, sin B, sin C` are in AP
Correct Answer is `=>` (C) AP

Radius of Circumcircle, inscribed circle, escribed circle

Q 2410045819

The angle subtended at the centre of a circle of
radius `3 ` cm by an arc of length `1` cm is
NDA Paper 1 2012
(A)

`(30^0)/pi`

(B)

`(60^0)/pi`

(C)

`60^0`

(D)

None of these

Solution:

`because theta = l/r = 1/3 = (1/3)^0xx(180^0)/pi = 60^0/pi`
Correct Answer is `=>` (B) `(60^0)/pi`
Q 2470423316

Consider a circle of radius `R`. What is the length
of a chord which subtends an angle `theta` at the
centre?
NDA Paper 1 2007
(A)

`2R sin(theta/2)`

(B)

`2R sin theta`

(C)

`2R tan (theta/2)`

(D)

`2R tan theta`

Solution:

In `Delta AOD`

`sin\ \ theta/2=(AD)/(OA)`

`=> sin\ \ theta/2=(AD)/R`

`=> AD=R sin \ \ theta/2`

`:.` Length of chord `AB = 2AD = 2R sin \ \ theta/2`
Correct Answer is `=>` (A) `2R sin(theta/2)`
Q 2542212133

Let p, q and r be the sides opposite to the angles P, Q and R, respectively in a `Delta PQR`. If `r^2 sin P sin Q = pq`, then the triangle is
WBJEE 2012
(A)

equilateral

(B)

acute angled but not equilateral

(C)

obtuse angled

(D)

right angled

Solution:

We know that in `Delta ABC`

`a/(sin A) = b/(sin B) = c/(sin C) = 2R`

`:. r^2 sin P sin Q = pq`

`=> r^2 · p/(2R_1) · q/(2 R_1) = pq`, where `R_1` is circumradius

of `Delta PQR`.

`=> r^2 = 4R_1^2`

`=> r = 2R_1`

` => 2R_1 sin R = 2R_1`

`=> R_1 = 90°`

`:. Delta PQR` is right angled.
Correct Answer is `=>` (D) right angled
Q 1580578417

The circumradius of the triangle whose sides are `13, 12` and `5`, is :
BITSAT 2006
(A)

`15`

(B)

`13/2`

(C)

`15/2`

(D)

`6`

Solution:

Let sides are `a= 13, b = 12, c = 5`

Now, `a^2 = b^2 + c^2`

`=> (13)^2 = (12)^2 + 5^2`

`=> 169 = 169`

`=> angleA = 90^0`

we know, `R = a/(2sinA)`

`R = 13/(2sin90^0) =13/2`
Correct Answer is `=>` (B) `13/2`
Q 1550723614

In a triangle, if ` r_1 = 2r_2 =3r_3` , then `a/b + b/c + c/a` is equal to
BITSAT 2008
(A)

`170/60`

(B)

`181/60`

(C)

`190/60`

(D)

`191/60`

Solution:

Given that, `r_1=2r_2=3r_3`

`Delta/(s-a) = (2 Delta)/(s-b) =( 3 Delta)/(s-c) = Delta/k`

Then , `s – a = k, s – b = 2 k, s – c = 3 k`

`3 s – (a + b + c) = 6 k` ,

`s = 6 k`

`a/5 = b/4 = c/3 =k`

Now, `a/b + b/c +c/a =5/4 + 4/3 + 3/5`

`(75+80+36)/60` = `191/60`
Correct Answer is `=>` (D) `191/60`

Area of triangle

Q 2379291116

If the angles of a triangle are `30°, 45°` and the
included side is `(sqrt 3 + 1)`, then what is the area of
the triangle?
NDA Paper 1 2013
(A)

`((sqrt 3+1)/2) cm^2`

(B)

`2 (sqrt 3+1) cm^2`

(C)

`((sqrt 3+1)/3) cm^2`

(D)

`((sqrt 3-1)/2) cm^2`

Solution:

Let `angle A = 30°, angle B = 45°` and `AB = sqrt 3 + 1`

Then, `angle C = 180°- (angle A+ angle B)`

(since, the sum of internal angles of a triangle is `180°`)

`=> angle C = 180°- (30° + 45°) = 180°- 75°= 105^o`

Now, by sine rule,

`(sin 30^o)/(BC)=( sin 105^o)/(sqrt 3 + 1)`

`:. BC=(sqrt 3+1) xx ((2 sqrt 2)/(sqrt 3+1)) xx 1/2=sqrt 2`

`[ sin 105^o=sin (60^o +45^o)=sin 60^o * cos 45^o+cos 60^o * sin 45^o
=sqrt 3/2 8 1/sqrt 2 +1/2 * 1/sqrt 2=(sqrt 3+1)/(2 sqrt 2)]`

Again, now by sine rule,

`(sin 45^o)/(AC)=(sin 105^o)/(sqrt 3+1)`

`=> AC=((sqrt 3+1))/(sqrt 2) xx (2 sqrt 2)/((sqrt 3+1))=2`

`:.` Area of `Delta ABC=1/2 xx BC xx AC xx sin 105^o`

`=1/2 xx 2 xx sqrt 2 xx ((sqrt 3+1))/(2 sqrt 2)=((sqrt 3+1)/2) cm^2`
Correct Answer is `=>` (A) `((sqrt 3+1)/2) cm^2`
Q 2460112915

For finding the area of a `triangle ABC`, which of the
following entities are required'?
NDA Paper 1 2009
(A)

Angles `A,B` and side a

(B)

Angles `A, B` and side b

(C)

Angles `A, B` and side c

(D)

Either (a) or (b) or (c)

Solution:

We know that, area of `triangle ABC` whose sides are `a, b` and
`c` are

`Delta=(c^2 sin A * sin B)/(2 sin C)=(a^2 sin B * sin C)/(2 sin A)=(b^2 sin C * sin A)/(2 sin B)`

(where, `A+ B+ C = 180^o`)

So, finding the area of `Delta ABC`, angles `A, B` and side `c` are required.
Correct Answer is `=>` (C) Angles `A, B` and side c
Q 1629223111

The area of the largest triangle that can be inscribed in a semi-circle of radius `r` is
CDS 2015
(A)

`r^2`

(B)

`2r^2`

(C)

`3 r^2`

(D)

`4 r^2`

Solution:

Let the `r` be the radius of semi-circle.

In `Delta ABC, AO = OB = OC = r`

`:.` Area of triangle `=1/2 xx` Base x Height

`=1/2 xx AB xx OC`

`=1/2 xx 2 r xx r =r^2`

Hence, the area of the largest triangle is `r^ 2`
Correct Answer is `=>` (A) `r^2`
 
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