Question

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

Answers
7 / (sqrt 13)
(sqrt 13) / 6
6 / (sqrt 13)
(sqrt 13) / 7

Answer 1

$\sin \theta =\dfrac{6}{7}\\ \cot \theta=\dfrac{x}{6}\\ x^2+6^2=7^2\\ x^2=49-36\\ x^2=13\\ x=\sqrt{13}\\ \boxed{\cot \theta =\dfrac{\sqrt{13}}{6}}$

Answer 2

Consider right triangle ABC with leg BC=6, hypotenuse AB=7 and angle C such that $\sin \angle A=\dfrac{6}{7}.$

By the Pythagorean theorem, you can find the length of the second leg AC:

$AC^2+BC^2=AB^2,\\AC^2+36=49,\\AC^2=13,\\AC=\sqrt{13}.$

Use the definition of $\cot$:

$\cot \angle A=\dfrac{\text{adjacent leg}}{\text{opposite leg}}=\dfrac{AC}{BC} =\dfrac{\sqrt{13}}{6}.$

Answer: $\cot \theta=\dfrac{\sqrt{13}}{6}.$