An acute angle θ is in a right triangle with sin θ = . what is the value of cot θ?

Asked by maham237 @ in Mathematics viewed by 281 People

Answered by maham237 @


Answer:

cot\theta=\frac{\sqrt{13}}{6}

Step-by-step explanation:

Step 1

Find the value of the cos\theta

we have that

sin\theta=\frac{6}{7}

Remember that

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

Substitute the value of sin\theta

\frac{6}{7}^{2}+cos^{2} \theta=1

\frac{36}{49}+cos^{2} \theta=1

cos^{2} \theta=1-\frac{36}{49}

cos^{2} \theta=\frac{13}{49}

square root both sides

cos \theta=\frac{\sqrt{13}}{7}

Step 2

Find the value of cot\theta

we know that

cot\theta=\frac{cos\theta}{sin\theta}

substitute the values

cot\theta=\frac{\frac{\sqrt{13}}{7}}{\frac{6}{7}}=\frac{\sqrt{13}}{6}


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