Question

If tanA=13/27,tanB=7/20 and A,BSNL are acute, show thatA+B=45​

Answer 1

Step-by-step explanation:

If tan A =13/27

tan B = 7/20 and A, B are acute, show that A + B = 45°

As we know that

$tan \:( A + \: B )= \Big( \frac{tan A+ tan B}{1 - tan A\:tan B}\Big) \\ \\ tan \: A = \frac{13}{27} \\ \\ tan \: B = \frac{7}{20} \\ \\ tan \:( A+ B) = \Bigg( \frac{ \frac{13}{27} + \frac{7}{20} }{1 - \frac{13}{27} \frac{7}{20} }\Bigg) \\ \\ = \Bigg (\frac{ \frac{260 +189 }{540} }{ \frac{540 - 91}{540} } \Bigg) \\ \\ = \Bigg( \frac{449}{449} \Bigg) \\ \\ = 1$

Thus

$tan (A+ B) = 1\\\\so\: (A+B)=tan^{-1}(1)\\\\(A+B)=tan^{-1}(tan 45°)\\\\so\\</p><p>A+B= 45°$

Hope it helps you.

Answer 2

Answer:

7 by 20 ×10

= i think 70 by 200