Question

sin(a+b)= root 3/2
cos(a-b)= 1/2​

Answer 1

Step-by-step explanation:

Answer

Given sin(A+B)=1

⇒sin(A+B)=sin90o (∵sin90o=1)

⇒A+B=90o …………(1)

Again, cos(A−B)=3/2

⇒cos(A−B)=cos30o (∵cos30o=3/2)

⇒A−B=30o …………(2)

Adding (1)+(2)

A+B+A−B=90o+30o

⇒2A=120o

⇒A=120/2

⇒A=60o

Putting A=60o in equation-(2) we get

A−B=30o

⇒60o−B=30o

⇒60o−30o=B

Answer 2

Given :-

• $sin(a + b) = \frac{ \sqrt{3} }{2}$
• $cos(a - b) = \frac{1}{2}$

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To Find :- Value of a and b.

$\huge\mathfrak{\color{red}{\underline{\underline{Solution♡}}}}$

From given,

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

$\sin(a + b) = \sin60$

Cancelling sin from both sides

a+b = 60° ------------------- (1)

Form given,

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

$\cos(a - b) = \cos60$

canceling cos from both sides

a–b = 60° ----------(2)

Adding (1) and (2)

2a = 120°

$a \: = \frac{120}{2} \\ \\ a \: = 60$

Putting value of a in (1) ,we get

60°+b = 60°

b = 60°–60°

b = 0°

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Hence, $\small{\bf{\blue{\fbox{\underline{\color{red}{a=60°}}}}}}$

$\small{\bf{\blue{\fbox{\underline{\color{blue}{b=0°}}}}}}$