Angle sum property of a triangle - class-VIII

Given 

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

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 $\triangle ABC$,
$\angle A - \angle B = 30$    ....(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

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

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 interior opposite angles

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

Remember this..

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

In $\triangle  ABC$

$\angle A+\angle B={ 90 }^{ \circ  }\ \angle B={ 90 }^{ \circ  }-\angle A$

Let A=(1, -2), B=(-7, 6), C=(11/5, 2/5). Vector AB = (-8, 8), vector AC = (11/5 - 1, 2/5 + 2) = (6/5, 12/5). The dot product AB dot AC = (-8)(6/5) + (8)(12/5) = -48/5 + 96/5 = 48/5. The magnitudes are |AB| = sqrt(64+64) = 8*sqrt(2) and |AC| = sqrt(36/25 + 144/25) = sqrt(180/25) = (6/5)sqrt(5). Cos(A) = (48/5) / (8sqrt(2) * (6/5)sqrt(5)) = 48 / (48sqrt(10)) = 1/sqrt(10). Since cos(A) is positive and approximately 0.316, A is acute and arccos(0.316) is approximately 71.5 degrees.

In $ \triangle ABC$

$\cfrac { x _{ 1 }+x _{ 2 } }{ 2 } =5\Rightarrow x _{ 1 }+x _{ 2 }=10.....(1)\quad and\quad y _{ 1 }+y _{ 2 }=12.....(2)$

The sum of angles in a triangle is 180 degrees. x + (x + 10) + (x + 20) = 180. 3x + 30 = 180, so 3x = 150, x = 50. The angles are 50, 60, and 70 degrees. The biggest is 70 degrees.

$\angle A + \angle B = 144$...(I)
$\angle A + \angle C = 124$...(II)
In triangle ABC,
$\angle A  + \angle B + \angle C = 180 $
Add, I and II,
$\angle A + \angle B + \angle A + \angle C = 144+ 124$
$180 + \angle A = 268 $
$\angle A = 268 - 180 $
$\angle A = 88$
Put this value in (I)
$\angle A + \angle B = 144$
$88 + \angle B = 144$
$\angle B = 56$
Put this value in (II)
$\angle A + \angle C = 124$
$88 + \angle C = 124$
$\angle C = 36$

The ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding sides.