$\cos 2\theta=2\cos^2 \theta-1=\frac{12}{13}\\\\2\cos^{2} \theta=\frac{25}{13}\\\\\cos^{2} \theta=\frac{25}{26}\\\\\boxed{\cos \theta=\frac{5}{\sqrt{26}}}$

(Note I took the positive case since $\theta$ terminates in the first quadrant)

Using the Pythagorean identity,

$\sin^2 \theta+\cos^2 \theta=1\\\\\sin^2 \theta+\frac{25}{26}=1\\\\sin^2 \theta=\frac{1}{26}\\\\\boxed{\sin \theta=\frac{1}{\sqrt{26}}}$

(Note I took the positive case since $\theta$ terminates in the first quadrant)

6 0

8 months ago

What is the length of line segment A.B?<br> Please help!!

weqwewe [10]

Answer:

x = 6.40 units

Step-by-step explanation:

4^2 + 5^2 = x^2

x = 6.40 units

5 0

2 years ago

Suppose you needed $ 22 comma 000$22,000 to maintain a particular standard of living in 19781978. How much would you have needed

wlad13 [49]

Sorry i dont know this is kind of tricky

5 0

2 years ago

M angle PQS= 3x; m angle SQR=5x-20 <br> Find x

Dahasolnce [82]

Let as consider the complete ques is : Ray QS bisects ∠PQR. Solve for x and find m∠PQR. m∠PQS = 3x ; m∠SQR = 5x-20.

Given:

Ray QS bisects ∠PQR.

m∠PQS = 3x

m∠SQR = 5x-20

To find:

The value of x and m∠PQR

Solution:

Ray QS bisects ∠PQR. So,

$m\angle PQS=m\angle SQR$

$3x=5x-20$

$3x-5x=-20$

$-2x=-20$

$x=10$

The value of x is 10.

Now,

$m\angle PQR=m\angle PQS+m\angle SQR$

$m\angle PQR=3x+5x-20$

$m\angle PQR=8x-20$

Put x=10,

$m\angle PQR=8(10)-20$

$m\angle PQR=80-20$

$m\angle PQR=60$

Therefore, the m∠PQR is 60 degrees.

5 0

3 years ago