Question

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Answer 1

hope this explanation helps you

Answer 2

`\textsf{b)} \quad -(3√(13))/(13)`

Explanation:

Trigonometric ratios

`\sf \sin(\theta)=(O)/(H)\quad\cos(\theta)=(A)/(H)\quad\tan(\theta)=(O)/(A)`

where:

• θ is the angle.
• O is the side opposite the angle.
• A is the side adjacent the angle.
• H is the hypotenuse (the side opposite the right angle).

`\textsf{If}\; \tan(\theta)=-(2)/(3) \implies \sf O=2 \; \textsf{and} \; A=3.`

Use Pythagoras Theorem to calculate the length of the hypotenuse:

`\sf \implies O^2+A^2=H^2`

`\implies \sf H=√(O^2+A^2)`

`\implies \textsf{H}=√(2^2+3^2)`

`\implies \textsf{H}=√(13)`

Therefore, substituting the values of A and H into the cosine ratio:

`\implies \cos \theta=(3)/(√(13))`

`\implies \cos \theta=(3)/(√(13))\cdot (√(13))/(√(13))`

`\implies \cos \theta=(3√(13))/(13)`

As the terminal side of the angle is in Quadrant II, and cosine is negative in Quadrants II and III, the exact value of cos(θ) is:

`\implies \cos \theta=-(3√(13))/(13)`