Tag: theorem of remote interior angles of a triangle

Questions Related to theorem of remote interior angles of a triangle

In $\triangle ABC$, if $b\cos A=a\cos B$ then the triangle is 

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. right angled

  2. isosceless

  3. equilateral

  4. scalene

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Correct Option: B
Explanation:

Given 

In $\triangle ABC,b\cos A=a\cos B$
$2{R}\sin B\cos A=2{R}\sin A \cos B$
$\sin (B-A)=0\implies B=A$

State true or false.
The sum of interior angles of a triangle is ${ 180 }^{ \circ  }$.

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. True

  2. False

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Correct Option: A
Explanation:

It is true that the sum of interior angles of a triangle is ${ 180 }^{ \circ  }$.

The angles of a triangle are in the ratio 2: 1: 3. Is the triangle right-angled triangle,

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. True

  2. False

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Correct Option: A
Explanation:

The angles of the triangle are in the ratio, 2:1: 3
Let the angles be $2x, x and 3x$
Thus, sum of the angles = 180
$2x + x+ 3x = 180$
$6x = 180$
$x = 30$ 
Hence, the angles will be 30, 60 and 90
Since, one of the angles is 90, the triangle is a right angled triangle.

In a $\triangle ABC$, $\angle A - \angle B = 30^{\circ}$ and $ \angle B -\angle C = 42^{\circ}$; find $\angle A$.

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $84^o$

  2. $94^o$

  3. $32^o$

  4. none of the above

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Correct Option: B
Explanation:

In $\triangle ABC$,
$\angle A - \angle B = 30$    ....(I)
$\angle B - \angle C = 42$     .....(II)
Also, sum of angles of the triangle $= 180$
$\angle A + \angle B + \angle C = 180$     ....(III)
On subtracting (I) and (II), we get

$\angle A-2\angle B+\angle C=-12^o$     ....(IV)
On subtracting (III) and (IV), we have
$3\angle B=192$
$\angle B=64^o$
From (I), we get
$\angle A=94^o$

If the angles of a triangle are in the ratio 2:3:4, find the three angles.

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $80^o, 120^o, 160^o$

  2. $20^o, 30^o, 40^o$

  3. $40^o, 60^o, 80^o$

  4. None of these

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Correct Option: C
Explanation:

The angles of a triangle are in the ratio 2:3:4 
Let $x:y:z=2:3:4$
Then $x= 2t; y= 3t; z= 4t$

Sum of all angles of a triangles is $ 180^0$.
$ 2t + 3t + 4t = 180^0 $
$ 9t = 180^0 $
$  t  = 20^0 $
$ x= 2\times 20^0= 40^0; y= 3\times 20^0= 60^0; z = 4\times 20^0= 80^0 $

In a $\triangle ABC$, the sides AB and AC have been produced to D and E. Bisectors of $\angle CBD$ and $\angle BCE$ meet at O. If $\angle A={ 64 }^{ 0 }$, then $\angle BOC$ is 

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. ${ 52 }^{ 0 }$

  2. ${ 58 }^{ 0 }$

  3. ${ 26 }^{ 0 }$

  4. ${ 112 }^{ 0 }$

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Correct Option: B
Explanation:

Given: OB and OC bisect $ext. \angle B$ and $ext. \angle C$, $\angle A = 64^{\circ}$

Now, In $\triangle OBC$,
Sum of angles = 180
$\angle OBC + \angle OCB + \angle BOC = 180$
$\frac{1}{2} (ext. \angle B + ext. \angle C) + \angle BOC = 180$ (OB and OC bisect exterior angles)
$\frac{1}{2} (180 - \angle ABC + 180 - \angle ACB) + \angle BOC = 180$
$\frac{1}{2} (360 - (\angle ABC + \angle ACB)) + \angle BOC = 180$
$\frac{1}{2} (360 - (180 - \angle A)) + \angle BOC = 180$ (Angle sum property)
$\frac{1}{2} (180 + \angle A) + \angle BOC = 180$
$\angle BOC = 180 - 90 -\frac{1}{2} (64)$
$\angle BOC = 58^{\circ}$

An exterior angle of a triangle is equal to the sum of two ______ opposite angles.

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. interior

  2. exterior

  3. vertical

  4. none of these

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Correct Option: A
Explanation:

An exterior angle of a triangle is equal to the sum of two interior opposite angles

In $\displaystyle \triangle ABC,\angle C=30^{\circ},\angle B=90^{\circ},BC=10 cm,BD\perp AC$ then the length of AD is

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $\displaystyle \frac{5}{\sqrt{3}}$ cm

  2. $\displaystyle \frac{6}{\sqrt{3}}$ cm

  3. $\displaystyle \frac{7}{\sqrt{3}}$ cm

  4. $\displaystyle \frac{8}{\sqrt{3}}$ cm

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Correct Option: A
Explanation:

In triangle ABC $\angle C=30^{0}and \angle B =90^{0}$ and BC=10 cm and $BD\perp AC$ 

Then $Sin 30^{0}=\frac{AB}{BC}\Rightarrow \frac{1}{2}=\frac{AB}{10}\Rightarrow AB=5 cm$
And $\angle ABD =60^{0}$
Then $tan 60^{0}=\frac{AB}{AD}\Rightarrow \sqrt{3}=\frac{5}{AD}\Rightarrow AD=\frac{5}{\sqrt{3}}cm$

The interior and boundary of a triangle is called

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. exterior

  2. interior

  3. triangular region

  4. plane

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Correct Option: C
Explanation:

Remember this..

The interior and boundary of a triangle is called triangular region..

One of the exterior angle of a triangle is $ 105^0$ and the interior opposite angles are in the ratio 2 : 5 . Find the angles of the triangle.

maths theorems on triangles theorem of remote interior angles of a triangle use of properties of parallel lines angle sum property of a triangle

  1. $ 30^o ; 45^o ; 105^o$ 

  2. $ 45^o ; 45^o ; 90^o$ 

  3. $ 30^o ; 75^o ; 75^o$ 

  4. $ 60^o ; 30^o ; 90^o$ 

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Correct Option: C
Explanation:

We have the property that, in a triangle exterior angle is equal to the sum of interior opposite angles.


Given, the interior opposite angles to the exterior angle $105^\circ$ are in the ration $2:5$

$\therefore 2x+5x=105^o$

$7x=105^o$ $\implies x=15^o$

Therefore the interior opposite angles to the angle $105^o$ are $2x=2(15)=30^o$ and $5x=5(15)=75^o$

Let the third angle of the triangle be $C$
We have the sum of interior angles of a triangle is $180^o$

$\therefore 30^o+75^o+C=180^o$
$C=180^o-75^o-30^o=180^o-105^o$
$C=75^o$

Hence, the angles are $30^o,75^o,75^o$.