Mathematics previous year questions Of Properties of Triangles For NDA

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previous year questions Of Properties of Triangles For NDA

previous year questions Of Properties of Triangles

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 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 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 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 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`

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 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 2339191912

In a `Delta ABC`, if the angles `A, B` and `C` are in A.P, then
which one of the following is correct?
NDA Paper 1 2012

(A)

`c=a+b`

(B)

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

(C)

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

(D)

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

Solution:

Given that, the angles `A, B` and `C` are in AP

i.e., `2B=A+C`...................(i)

Also, in `Delta ABC` `A+B+C=180^o` (from (i))

`=>2B+B=180°`

`=>B = 60°`

`=>cos B=(a^2 + c^2- b^2)/(2ac)=cos 60^o`

`=>1/2=(a^2 + c^ 2 - b^ 2)/(2ac)`

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

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

Correct Answer is `=>` (D) `b^2=a^2+c^2-ac`

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 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 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 2460401315

If the sides a, `b, c` of a `Delta ABC` are in arithmetic
progression and `'a'` is the smallest side, then what
is`cos A` equal to?
NDA Paper 1 2011

(A)

`(3c-4b)/(2c)`

(B)

`(3c-4b)/(2c)`

(C)

`(4c-3b)/(2c)`

(D)

`(3b-4c)/(2c)`

Solution:

Given that, the sides `a, b` and `c` of a `Delta ABC` are in AP.

Then, `2b =a+ c`....................(i)

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

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

[from Eq. (i) `a= 2b- c => a2 = (2b - c)^2]`

`=(b^2+c^2-4b^2-c^2+4bc)/(2bc)`

`=(4bc-3b^2)/(2bc)=(4c-3b)/(2c)`

Correct Answer is `=>` (C) `(4c-3b)/(2c)`

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 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

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

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

Q 2460112015

`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 `tan(A- C)` equal to?

NDA Paper 1 2011

(A)

`0`

(B)

`1`

(C)

`2`

(D)

None of these

Correct Answer is `=>` (D) None of these

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 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 2450412314

If angles of a triangle are in the ratio `1 : 2 : 3`, then
what is the ratio of its corresponding sides'?
NDA Paper 1 2009

(A)

`3:2:1`

(B)

`1: sqrt 2 : sqrt 3`

(C)

`1: sqrt 3 :2`

(D)

`2: sqrt 3:4`

Solution:

Let the angles of a triangle be `x, 2x` and `3x`.

`:. x + 2x + 3x = 180°`

`=> x= 30°`

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

By using sine rule,

`a/(sin 30^o)=b/(sin 60^o)=c/(sin 90^o)`

`=> a/(1/2)=b/(sqrt 3/2)=c/1`

`=> a/1=b/sqrt 3=c/2`

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

Correct Answer is `=>` (C) `1: sqrt 3 :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 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 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 2410012819

If the sides of a triangle are `6 cm, 10 cm` and
`14 cm`, then what is the largest angle induded by
the sides?
NDA Paper 1 2009

(A)

`90^o`

(B)

`120^o`

(C)

`135^o`

(D)

`150^o`

Solution:

We know that, the largest side has the greatest angle
opposite it.

`:. a=6 , b=10 cm` and `c=14 cm`

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

`=(36 +100-196)/(2 xx 6 xx 10)`

`:. angle C=120^o`

Correct Answer is `=>` (B) `120^o`

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)

Correct Answer is `=>` (C) Angles `A, B` and side c

Q 2451023824

If the sines of two angles of a triangle are equal to
`5/13` and `99/101` then what is the cosine of the third
angle?
NDA Paper 1 2009

(A)

`255/1313`

(B)

`265/1313`

(C)

`275/1313`

(D)

`770/1313`

Solution:

Here `sintheta = 5/13` and `sinphi = 99/101`

`therefore cos{pi-(theta+phi)} = -cos(theta+phi)`

` = -(costheta cosphi-sinthetasinphi)`

` = -{sqrt(1-25/169) sqrt(1-(99/101)^2) - 5/13* 99/101}`

` = -(12/13 * 20/101 - 5/13 * 99/101) = -(240/1313-495/1313) = 255/1313`

Correct Answer is `=>` (A) `255/1313`

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 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 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 2430278112

If two angle so triangle are `tan ^(-1)( 1/2)` and `tan ^(-1)(1/3)`
then what is the third angle?
NDA Paper 1 2007

(A)

`30^o`

(B)

`45^o`

(C)

`90^o`

(D)

`135^o`

Solution:

We know that, in `triangle ABC, angle A + angle B + angle C = pi`

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

`=>tan^(-1) ((5//6)/(5//6)) +angle C=pi => pi/4+angle C=pi`

`:. angle C=pi-pi/4=(3pi)/4=135^o`

Correct Answer is `=>` (D) `135^o`

Q 2480623517

In a `triangle ABC`, if `a = 2b` and `A = 3B`, then which one of
the following is correct?
NDA Paper 1 2007

(A)

The triangle is obtuse angled

(B)

The triangle is acute angled but not right angled

(C)

The triangle is right angled

(D)

The triangle is isosceles, but not obtuse angled

Solution:

We know that

`a/(sin A) =b/(sin B)` (by sine rule)

`=> (2b)/(sin 3B)=b/(sin B)` `(a=2b and A=3B)`

`=> 2 sin B = sin 3B`

`=> 2 sin B = 3 sin B - 4 sin^3 B` `(·: sin ^3x = 3 sinx-4 sin^3 x)`

`=> 0 = sin B - 4 sin^3 B`

`=> 0 = sin B (1 - 4 sin^2 B)`

`=> sin B = 0` or `1 - 4 sin^2 B = 0`

when `sin B = 0`, then `B = 0^o`

and `A=3B=0°`

which is not possible.

and when `1 - 4 sin^2 B = 0 => sin B = ± 1/2`

`=> B = 30°` and `A = 90°`

So, the triangle is right angled triangle

`

Correct Answer is `=>` (C) The triangle is right angled