Question

please help me to solve this question as soon as possible please help guys​

Answer 1

Answer:

Question no 2

answer is

cos (40+x)° = sin30°

sin {90-(40+x) }= sin 30°

sin(50-x)° = sin30°

so,

50-x =30

-x = 30-50

-x =-20

hence ,

x = 20°

Answer 2

Answers:

Answer of Q.43 is x = $\frac{31}{8}$

Answer of Q.43 is x = 20°

Step-by-step explanation:

Given Problem Q.43:

If x $cot^2$ 45 - $sec^2$ 60+$sin^2$  30 =$\frac{1}{8}$ ,then find the value of x

Solution:

To Find:

Value of x

-------------------

Method:

x - 4 +$\frac{1}{4}$ = $\frac{1}{8}$

As,

Cot 45 = 1 ;sec 60 = 2; sin 30 = $\frac{1}{2}$

x = $\frac{31}{8}$

Given Problem Q.44

If cos (40˚ + x) = sin 30˚, find the value of x

Solution:

To Find:

Value of x

----------------

Method:

As we know,

Sin(90°-α) = cosα

∴Sin30° = sin(90°- 60°)

=cos60°

∵cos(40°+ x) = sin 30°

⇒cos(40°+ x) = cos 60°

⇒40°+ x = 60°

⇒x = 20°