Question

if cos ( a + 4 b) = 1/ 2 and sin( a+2b)= 1/✓ 2 then find the value of a and b where a> b​

Answer 1

EXPLANATION.

→ Cos ( a + 4b ) = 1/2

→ Sin ( a + 2b ) = 1/√2

→ To find value of a and b.

→ Cos ( a + 4b ) = Cos 60° ......(1)

→ Sin ( a + 2b ) = Sin 45°. ......(2)

→ a + 4b = 60° .......(1).

→ a + 2b = 45° .......(2).

→ From equation (1) and (2) we get,

→ 2b = 15°.

→ b = 15/2

→ Put the value of b = 15/2 in equation (1)

we get,

→ a + 4 X 15/2 = 60°.

→ a + 30° = 60°.

→ a = 30°.

→ Value of a = 30° and b = 15/2°.

Answer 2

Question :

If cos (a + 4b) = 1/2 and sin(a + 2b) = 1/√ 2 then find the value of a and b where a > b

Given that :

• cos (a + 4b) = 1/2
• sin(a + 2b) = 1/√2
• a>b

To find :

• Value of a and b where a>b

Solution :

• Value of a = 30°
• Value of b = 15/2°

Full solution :

As we know that is will be written like this

Before :

• cos (a + 4b) = 1/2
• sin(a + 2b) = 1/√2

After :

• cos (a + 4b) = cos 60° ᗴǫᴜᴀᴛɪᴏɴ⑴

• sin(a + 2b) = sin 45° ᗴǫᴜᴀᴛɪᴏɴ⑵

Now,

• cos (a + 4b) = cos 60°

Cos and Cos cut each other hence,

• (a+4b) = 60° ᗴǫᴜᴀᴛɪᴏɴ⑴

• sin(a + 2b) = sin 45°

Sin and Sin cut each other hence,

• (a+2b) = 45° ᗴǫᴜᴀᴛɪᴏɴ⑵

Now, from ᗴǫᴜᴀᴛɪᴏɴ⑵ and ᗴǫᴜᴀᴛɪᴏɴ⑴ we get the following results.

✧ (a+4b) = 60° - (a+2b) = 45°

• a and a cancel each other.

✧ 4b = 60° -2b = 45°

✧ 4b - 2b = 60° - 45°

✧ 2b = 15°

✧ b = 15/2

Now we have to substitute the value of b = 15/2 in the ᗴǫᴜᴀᴛɪᴏɴ⑴

✧ a + 4 × 15/2 = 60°

• Cancelling 4 by 2 we get 2 hence,

✧ a + 2 × 15 = 60°

✧ a + 30 = 60°

✧ a = 60 - 30

✧ a = 30°

• Hence, the value of a is 30° and b is 15/2°