Question

sin(A-B) =√3/2 and cos (A+B)=1/2 0°<A+B<90° find a and b
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Answer 1

$\LARGE{ \underline{\underline{ \sf{Required \: answer:}}}}$

GiveN:

• sin (A - B) = √3 / 2

This means,

⇒ sin (A - B) = sin 60°

⇒ A - B = 60°

⇒ A = B + 60°

AnD,

• cos (A + B) = 1/2

This means,

⇒ cos(A + B) = cos 60°

⇒ A + B = 60°

Plugging B + 60° in place of A,

⇒ B + 60° + B = 60°

⇒ 2B = 0°

⇒ B = 0°

Then, A = 60°

Hence,

The required measures of Angle A and B are:

$\huge{ \boxed{ \sf{ \purple{60 \degree \: and \: 0 \degree}}}}$

Note that!!

• The measure of A and B is not less greater than 0°, It is their sum A + B that is greater than 0° and less/equal to 90°.

• The above condition is being satisfied!

Answer 2

Answer:

$\huge \tt \: Solution$

$\sin(a - b) = \frac{ \sqrt{3} }{2}$

$\sin(a - b) = sin 60 \degree$

$a \: - b \: = 60 \degree$

$a \: = b \: + 60$

Then,

$\cos(a + b) = \frac{1}{2}$

$\cos(a + b) = \cos(60 \degree)$

$a \: + b \: = 60 \degree$

Now,

Plugging B + 60⁰ in place of A

B + 60⁰+ B = 60⁰

2B = 0⁰

B = 0⁰

A = 60⁰

$\huge \fbox {0 \degree \: and \: 60 \degree}$