Question

Iif tan(A+B)+1 and cos(A-B)=$\sqrt{3}$/2, A>B then find A and B

Answer 1

Question:

If tan(A+B) = 1 and cos(A-B) = √3/2, A > B, then find A and B.

Answer:

A = 37.5°

B = 7.5°

Solution:

Given:

tan(A+B) = 1 --------(1)

cos(A-B) = √3/2 -------(2)

Considering eq-1 ,we have;

=> tan(A+B) = 1

=> tan(A+B) = tan45°

=> A+B = 45° -------(3)

Considering eq-2 ,we have;

=> cos(A-B) = √3/2

=> cos(A-B) = cos30°

=> A-B = 30° ----------(4)

Adding eq-3 and eq-4 , we get;

=> A+B+A-B = 45° + 30°

=> 2A = 75°

=> A = 75°/2

=> A = 37.5°

Now, putting A = 37.5° in eq-4 , we get;

=> A-B = 30°

=> 37.5° - B = 30°

=> B = 37.5° - 30°

=> B = 7.5°

Hence,

Required values of A and B are 37.5° and 7.5° respectively.

Answer 2

$\huge{\underline{\underline{\red{\mathbf{Answer :}}}}}$

$\Large{\underline{\bf{Given :}}}$

tan(A + B) = 1

Cos(A - B) = √3/2

_______________________

$\Large{\underline{\bf{To \: Find :}}}$

Value of A and B

_______________________

$\Large{\underline{\bf{Solution :}}}$

We know that,

$\large{\boxed{\purple{\sf{tan \: 45^{\circ} = 1}}}}$

So,

$\sf{ \cancel{tan}(a \: + \: b) = \cancel{tan }\: 45^{ \circ} }$

A + B = 45° .......(1)

_____________________

Now,

$\large{\boxed{\blue{\sf{Cos \: 30^{\circ} = \frac{\sqrt{3}}{2}}}}}$

So,

$\sf{ \cancel{cos}(a \: - \: b) = \cancel{cos} \: {30}^{ \circ} }$

A - B = 30° .......(2)

____________________

★ From equation 1.

A = 45° - B

★ Put Value of A in equation 2.

45° - B - B = 30°

45° - 2B = 30°

-2B = 30° - 45°

-2B = -15°

B = -15°/-2

B = 7.5°

$\large{\boxed{\orange{\sf{B = 7.5^{\circ} }}}}$

____________________

★ Put Value of B in equation 1.

A + 7.5° = 45°

A = 45° - 7.5°

A = 37.5°

$\large{\boxed{\green{\sf{A = 37.5^{\circ} }}}}$

$\rule{200}{2}$

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