Question

ܛܠܪ
find the value of A and B
tam (A+B) = 3 and
tam (A-B) = 1/underoot3
​

Answer 1

Tigers are preserved in Corbett National park

Answer 2

$\large \boxed{\large \star \: \mathtt{answer : }}$

$\sf{tan(A+B) = \sqrt{3} } \\ \\ \sf{tan(A - B) = \frac{1}{ \sqrt{3} } }$

$\sf{we \: know : } \\ \sf{tan60 \degree = \sqrt{3} } \\ \sf{tan(A + B) = \sqrt{3} } \\ \sf{tan(A+B) = tan60 \degree} \\ \sf{A+B = 60 \degree \: \: \: \: \: \: \: \: \: \: \: \: ...(i)}$

$\sf{we \: know : } \\ \sf{tan30 \degree = \frac{1}{ \sqrt{3} } } \\ \sf{tan(A - B) = \frac{1}{ \sqrt{3}}} \\ \sf{tan(A - B) = tan30 \degree} \\ \sf{A - B =30 \degree \: \: \: \: \: \: \: \: ...(ii)}$

$\sf{by \: equation \: (i) \: and \: (ii)} \\ \sf{A+ \cancel{B }= 60 \degree} \\ \sf{ + A - \cancel{B} = 30 \degree } \\ \sf{2 A= 90 \degree } \\ \boxed{ \sf{A = 45 \degree}} \\ \sf{A +B = 60 \degree } \\ \sf{45 \degree + B = 60 \degree} \\ \sf{B = 60 \degree - 45 \degree} \\ \boxed{\sf{B = 15 \degree}}$