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2. Find the values of the unknowns x and y in the following diagrams

Answers (1)

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i) As we know, the exterior angle is equal to the sum of opposite internal angles in a triangle.

$50^0+x=120^0$

$x=120^0-50^0$

$x=70^0$

Now, As we know the sum of internal angles of a triangle is 180. so,

$50^0+x+y=180^0$

$50^0+70^0+y=180^0$

$y=180^0-50^0-70^0$

$y=60^0$

Hence, $x=70^0\:\:and\:\:y=60^0$ .

ii) As we know when two lines are intersecting, the opposite angles are equal. So

$y=80^0$

Now, As we know the sum of internal angles of a triangle is 180. so,

$50^0+y+x=180^0$

$50^0+80^0+x=180^0$

$x=180^0-50^0-80^0$

$x=50^0$

Hence, $x=50^0\:\:and\:\:y=80^0$ .

iii) As we know, the exterior angle is equal to the sum of opposite internal angles in a triangle

$x=50^0+60^0$

$x=110^0$

Now, As we know the sum of internal angles of a triangle is 180. so,

$y+50^0+60^0=180^0$

$y=180^0-50^0-60^0$

$y=70^0$

Hence, $x=110^0\:\:and\:\:y=70^0$ .

iv) As we know when two lines are intersecting, the opposite angles are equal. So

$x=60^0$

Now, As we know the sum of internal angles of a triangle is 180. so,

$30^0+y+x=180^0$

$30^0+y+60^0=180^0$

$y=180^0-30^0-60^0$

$y=90^0$

Hence, $x=60^0\:\:and\:\:y=90^0$

v) As we know when two lines are intersecting, the opposite angles are equal. So

$y=90^0$

Now, As we know the sum of internal angles of a triangle is 180. so,

$y+x+x=180^0$

$90^0+2x=180^0$

$2x=90^0$

$x=45^0$

Hence, $x=45^0\:\:and\:\:y=90^0$

vi)As we know when two lines are intersecting, the opposite angles are equal. So

$y=x$

Now, As we know the sum of internal angles of a triangle is 180. so,

$x+x+x=180^0$

$3x=180^0$

$x=60^0$

Hence, $x=60^0\:\:and\:\:y=60^0$ .

Posted by

Pankaj Sanodiya

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