Question

ABCD is a square DM is 3cm and BN is 4 cm and angle MAN is 45 degree find the value of mn​

Answer 1

Step-by-step explanation:

Given ABCD is a square DM is 3cm and BN is 4 cm and angle MAN is 45 degree find the value of mn

• Now 3 + p = 4 + q  
•     Or p – q = 1
•   Or  q = p – 1
• So tan (45 – theta) = 3 / 3 + p  
• 1 – tan theta / 1 + tan theta  can be used
• Or 6 + p / - p = 2 / - 2 tan theta
• Or p / 6 + p = tan theta
• Now tan theta = 4 / 3 + p
• Or 4 / p + 3 = p / 6 + p
• Now 24 p + 4 p = p^2 + 3p
• Or p^2 – p – 24 = 0
• So p = - (-1) + - √(-1)^2 – 4 (1) (-24) / 2
•        = 1 + - √1 + 96 / 2
•        = 1 + √97 / 2 and  √97 – 1 / 2
• So MN^2 = p^2 + q^2  
• = (1 + √97)^2 + (√97 – 1)^2 / 4
• = 2 (1 + 97 ) / 4
• So MN^2 = 98 / 2 = 49
• Or MN = 7 cm