Question

if A is an obtuse angle whose sine is 5/13 and B is an acute angle whose tangent is 3/4 without using tables find values of sin2B and tan (A-B) ​

Answer 1

$\textbf{Given:}$

$sinA=\dfrac{5}{13}$

$tanB=\dfrac{3}{4}$

$\textbf{To find:}$

$\text{The values of sin2B and tan(A-B) }$

$\textbf{Solution:}$

$\text{Consider,}$

$cos^2A=1-sin^2A$

$cos^2A=1-\dfrac{25}{169}$

$cos^2A=\dfrac{144}{169}$

$cosA=\pm\dfrac{12}{13}$

$\text{Since A lies in second quadrant, we have}$

$cosA=\dfrac{-12}{13}$

$tanA=\dfrac{sinA}{cosA}$

$tanA=\dfrac{5/13}{-12/13}$

$tanA=\dfrac{-5}{12}$

$\text{Now,}$

$sin2B$

$=\dfrac{2\,tanB}{1+tan^2B}$

$=\dfrac{2(\frac{3}{4})}{1+\frac{9}{16}}$

$=\dfrac{\frac{6}{4}}{\frac{25}{16}}$

$=\dfrac{3}{2}{\times}\dfrac{16}{25}$

$=\dfrac{3}{1}{\times}\dfrac{8}{25}$

$=\dfrac{8}{25}$

$\implies\boxed{\bf\,sin2B=\dfrac{8}{25}}$

$tan(A-B)$

$=\dfrac{tanA-tanB}{1+tanA\,tanB}$

$=\dfrac{\frac{-5}{12}-\frac{3}{4}}{1+(\frac{-5}{12})(\frac{3}{4})}$

$=\dfrac{\frac{-5-9}{12}}{1+(\frac{-5}{4})(\frac{1}{4})}$

$=\dfrac{\frac{-14}{12}}{1-\frac{5}{16}}$

$=\dfrac{\frac{-7}{6}}{\frac{11}{16}}$

$=\dfrac{-7}{6}{\times}\dfrac{16}{11}$

$=\dfrac{-7}{3}{\times}\dfrac{8}{11}$

$=\dfrac{-56}{33}$

$\implies\boxed{\bf\,tan(A-B)=\dfrac{-56}{33}}$

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Answer 2

Step-by-step explanation:

Given if A is an obtuse angle whose sine is 5/13 and B is an acute angle whose tangent is 3/4 without using tables find values of sin 2B and tan (A-B) ​

• Given Sin A = 5/13 , tan B = ¾
• So we have by Pythagoras theorem, AC^2 = AB^2 + BC^2
• So tan B = ¾ = opposite / adjacent
• So AC^2 = 3^2 + 4^2
•                = 9 + 16
•                = 25
• Or AC = 5  
• So hypotenuse = 5
• Now we have Sin B = 3/5 = opposite / hypotenuse.
• Also cos B = 4/5 = Adjacent / hypotenuse.  
•  We have  Sin 2 B = 2 Sin B Cos B
•              = 2 x 3/5 x 4/5
•             = 24 / 25
• Now we have Sin A = 5/13
• So we can write Cos A = √1 – sin^2 A
•                = √1 – (5/13)^2
•               = √1 – 25 / 169
•              = √169 – 25 / 169
•                              = √144 / 169
•                             = 12 / 13
• Now Sin A = 5/13 and Cos A = 12/13
• Therefore we get tan A = Sin A / Cos A
•                          = 5/13 / 12 / 13
•                           = 5/12
• Also tan B = 3/4  
• So we have tan A – B = tan A – tan B / 1 + tan A tan B
•                                     = 5/12 – ¾ / 1 + 5/12 x ¾  
•                                      = 5 – 9 / 12 / 1 + 5 / 16
•          By taking LCM we get = - 4/12 / 16 + 5 / 16
•             = Simplifying we get = - 1/3 / 21 / 16
•                                        = - 16 / 63
•              So tan A – B = - 16 / 63

Reference link will be

https://brainly.in/question/17911896