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qaws [65]
3 years ago
8

4.) What is the exact value of sinθ when θ lies in Quadrant II and cosθ=−513

Mathematics
1 answer:
dimaraw [331]3 years ago
5 0

Answer:

Part 4) sin(\theta)=\frac{12}{13}

Part 10) The angle of elevation is 40.36\°

Part 11) The angle of depression is 78.61\°

Part 12) arcsin(0.5)=30\°  or arcsin(0.5)=150\°

Part 13) -45\°  or 225\°

Step-by-step explanation:

Part 4) we have that

cos(\theta)=-\frac{5}{13}

The angle theta lies in Quadrant II

so

The sine of angle theta is positive

Remember that

sin^{2}(\theta)+ cos^{2}(\theta)=1

substitute the given value

sin^{2}(\theta)+(-\frac{5}{13})^{2}=1

sin^{2}(\theta)+(\frac{25}{169})=1

sin^{2}(\theta)=1-(\frac{25}{169})  

sin^{2}(\theta)=(\frac{144}{169})

sin(\theta)=\frac{12}{13}

Part 10)

Let

\theta ----> angle of elevation

we know that

tan(\theta)=\frac{85}{100} ----> opposite side angle theta divided by adjacent side angle theta

\theta=arctan(\frac{85}{100})=40.36\°

Part 11)

Let

\theta ----> angle of depression

we know that

sin(\theta)=\frac{5,389-2,405}{3,044} ----> opposite side angle theta divided by hypotenuse

sin(\theta)=\frac{2,984}{3,044}

\theta=arcsin(\frac{2,984}{3,044})=78.61\°

Part 12) What is the exact value of arcsin(0.5)?

Remember that

sin(30\°)=0.5

therefore

arcsin(0.5) -----> has two solutions

arcsin(0.5)=30\° ----> I Quadrant

or

arcsin(0.5)=180\°-30\°=150\° ----> II Quadrant

Part 13) What is the exact value of arcsin(-\frac{\sqrt{2}}{2})

The sine is negative

so

The angle lies in Quadrant III or Quadrant IV

Remember that

sin(45\°)=\frac{\sqrt{2}}{2}

therefore

arcsin(-\frac{\sqrt{2}}{2}) ----> has two solutions

arcsin(-\frac{\sqrt{2}}{2})=-45\° ----> IV Quadrant

or

arcsin(-\frac{\sqrt{2}}{2})=180\°+45\°=225\° ----> III Quadrant

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