If $2 cos (A + B) = 2 sin (A - B) = 1$ ; find the value of $B$ in degrees

Open in App

Solution

$2 c o s (A + B) = 2 s i n (A - B) = 1$
$2 c o s (A + B) = 1$
$c o s (A + B) = \frac{1}{2}$
$c o s (A + B) = c o s 60$
$A + B = 60$ ...(I)
$2 s i n (A - B) = 1$
$s i n (A - B) = \frac{1}{2}$
$s i n (A - B) = s i n 60$
$A - B = 30$ ...(II)
Add I and II,
$2 A = 90$
$A = 45^{\circ}$
Subtract I and II,
$2 B = 30$
$B = 15^{\circ}$

Suggest Corrections

Similar questions

2 cos (A - B) = 2 sin (A + B) = \sqrt{3},

A

If 2 cos(A + B) = 1, and 2 sin(A –B) = 1 then the values of A and B are

\sqrt{2} cos A = cos B + {cos}^{3} B

\sqrt{2} sin A = sin B - {sin}^{3} B

| sin (A - B) |

∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} s i n \frac{A}{2} & s i n \frac{B}{2} & s i n \frac{c}{2} s i n (A + B + C) & s i n \frac{B}{2} & c o s \frac{A}{2} c o s (\frac{A + B + C}{2}) & t a n (A + B + C) & s i n \frac{C}{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

A, B, C

A B C

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} sin \frac{A}{2} & sin \frac{B}{2} & sin \frac{C}{2} sin (A + B + C) & sin \frac{B}{2} & cos \frac{A}{2} cos \frac{(A + B + C)}{2} & tan (A + B + C) & sin \frac{C}{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣

Join BYJU'S Learning Program