Given
⇢m∠JNM = (7x+2)°
⇢m∠KNP = (10x + 3)°
⇢m∠LNP = 90°
⇢m∠PNM = 90°
To find
⇢m∠ KNP
⇢m∠ LNK
⇢m∠ JNL
⇢m∠ JNP
Solution
In line KJ,
m∠ KNP + m∠ JNM + m∠PNM = 180°
- The measure of a straight line angle is 180°
putting the values given in the above equation,
⇢(10x + 3)° + (7x+2)° + 90° = 180°
⇢(10x+7x)° + (3+2 +90)° = 180°
⇢17x° + 95° = 180°
⇢17x° = 180°-95°
⇢17x° = 85°
⇢x = (85/17)°
⇢ x = 5°
hence,
⇢m∠ KNP = (10x + 3)° = (10 x 5 +3)° = 53°
⇢m∠ JNP = (7x+2)° + 90° = (7 x 5+2+90)° = 127°
⇢m∠ LNK = 180° - (90+53)° = 37°
⇢m∠ JNL = 180° - 37° = 143°
Given
⇢m∠Q= (31-3x)°
⇢m∠R= (19x-5)°
⇢m∠Q & ∠R are complementary angles
To find
⇢m∠R
Solution
Given that m∠Q & ∠R are complementary angles,
so ,
m∠Q + ∠R = 90°
- The sum of two complementary angles=90°
putting the values given in the above equation,
⇢(31-3x)° + (19x-5)° = 90°
⇢(19x-3x)° + (31-5)° = 90°
⇢16x° + 26° = 90°
⇢ 16x° = 90° -26°
⇢16x° = 64°
⇢ x = (64/16)°
⇢x = 4
so,
m∠R = (19x-5)° = (19 x 4 -5)° = 71°
Given
∠V & ∠W are supplementary angles
To find
The measure of ∠V & ∠W
Solution
Given that,
∠V = 6∠W - 2° _______(i)
Also,
∠V + ∠W = 180°_______(ii)
- The sum of two supplementary angles= 180°
putting the value of ∠V in equation (ii)
⇢6∠W - 2° + ∠W = 180°
⇢ 6∠W + ∠W - 2° = 180°
⇢7∠W -2° = 180°
⇢7 ∠W = 180°+ 2°
⇢7 ∠W = 182°
⇢ ∠W = 182°/7
⇢ ∠W = 26°
and
the value of ∠V
⇢= 180° -26°
⇢= 154°
hence , the value of ∠V = 154° & ∠W = 26°
