Suppose ABC is a right triangle with sides of lengths a, b, and c and right angle at C. Find the unknown side length using
the Pythagorean theorem and then find the value of the indicated trigonometric function of the given angle. Rationalize
the denominator if applicable.
-Find $\tan B$ when $b = 6$ and $c = 7$.
A) $\frac { 7 \sqrt { 13 } } { 13 }$

B) $\frac { \sqrt { 13 } } { 7 }$

C) $\frac { 6 \sqrt { 13 } } { 13 }$

D) $\frac { \sqrt { 13 } } { 6 }$

Question

Suppose ABC is a right triangle with sides of lengths a, b, and c and right angle at C. Find the unknown side length using
the Pythagorean theorem and then find the value of the indicated trigonometric function of the given angle. Rationalize
the denominator if applicable.
-Find $\tan B$ when $b = 6$ and $c = 7$.
A) $\frac { 7 \sqrt { 13 } } { 13 }$

B) $\frac { \sqrt { 13 } } { 7 }$

C) $\frac { 6 \sqrt { 13 } } { 13 }$

D) $\frac { \sqrt { 13 } } { 6 }$

Quizplus · Accepted Answer

First, use the Pythagorean theorem $a^2 + b^2 = c^2$ to find the unknown side $a$. Given $b = 6$ and $c = 7$, we have $a^2 + 6^2 = 7^2$, which simplifies to $a^2 + 36 = 49$. Solving for $a$, we find $a^2 = 13$, so $a = \sqrt{13}$. The tangent of angle B, $\tan B$, is the ratio of the opposite side to the adjacent side, which is $\frac{a}{b} = \frac{\sqrt{13}}{6}$. However, the correct answer is $\tan B = \frac{\text{opposite to angle B}}{\text{adjacent to angle B}} = \frac{6}{\sqrt{13}}$, which after rationalizing the denominator becomes $\frac{6\sqrt{13}}{13}$.

Suppose ABC Is a Right Triangle with Sides of Lengths tan⁡B\tan BtanB

Suppose ABC Is a Right Triangle with Sides of Lengths $\tan B$