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Flashcards: Tangent

Right

Refer to the triangle in the above diagram. Which of the following expressions correctly gives its area?

$\frac{72}{ \tan 35^{\circ }}$ $\displaystyle \frac{72}{ \tan 35^{\circ }}$

$\frac{144}{ \tan 35^{\circ }}$ $\displaystyle \frac{144}{ \tan 35^{\circ }}$

$144 \tan 35^{\circ }$ $\displaystyle 144 \tan 35^{\circ }$

$72 \tan 35^{\circ }$ $\displaystyle 72 \tan 35^{\circ }$

None of the other choices gives the correct response.

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Flashcards: Tangent

Right

Refer to the triangle in the above diagram. Which of the following expressions correctly gives its area?

$\frac{72}{ \tan 35^{\circ }}$ $\displaystyle \frac{72}{ \tan 35^{\circ }}$

$\frac{144}{ \tan 35^{\circ }}$ $\displaystyle \frac{144}{ \tan 35^{\circ }}$

$144 \tan 35^{\circ }$ $\displaystyle 144 \tan 35^{\circ }$

$72 \tan 35^{\circ }$ $\displaystyle 72 \tan 35^{\circ }$

None of the other choices gives the correct response.

Report an Error Copyright Notice

Flashcards: Tangent

Refer to the triangle in the above diagram. Which of the following expressions correctly gives its area?

Your answer:

$\frac{72}{ \tan 35^{\circ }}$ $\displaystyle \frac{72}{ \tan 35^{\circ }}$

Your answer:

$\frac{144}{ \tan 35^{\circ }}$ $\displaystyle \frac{144}{ \tan 35^{\circ }}$

Your answer:

$144 \tan 35^{\circ }$ $\displaystyle 144 \tan 35^{\circ }$

Your answer:

None of the other choices gives the correct response.

Correct answer:

$72 \tan 35^{\circ }$ $\displaystyle 72 \tan 35^{\circ }$

Explanation:

The area of a right triangle is half the product of the lengths of its legs, which here are $\overline{AB }$ $\displaystyle \overline{AB }$ and $\overline{AC}$ $\displaystyle \overline{AC}$ - that is,

$a= \frac{1}{2}(AB)(AC)$ $\displaystyle a= \frac{1}{2}(AB)(AC)$

We are given that AC=12 $\displaystyle AC=12$. $\overline{AB }$ $\displaystyle \overline{AB }$ is the leg opposite the angle $\angle C$ $\displaystyle \angle C$ and $\overline{AC}$ $\displaystyle \overline{AC}$ is its adjacent leg, we can find $\displaystyle AB$ using the tangent ratio:

$\tan (m \angle C) = \frac{AB}{AC}$ $\displaystyle \tan (m \angle C) = \frac{AB}{AC}$

Setting AC=12 $\displaystyle AC=12$ and $m \angle C = 35 ^{\circ }$ $\displaystyle m \angle C = 35 ^{\circ }$, we get

$\tan 35^{\circ } = \frac{AB}{12}$ $\displaystyle \tan 35^{\circ } = \frac{AB}{12}$

Solve for $\displaystyle AB$ by multiplying both sides by 12:

$12 \cdot \tan 35^{\circ } =12 \cdot \frac{AB}{12}$ $\displaystyle 12 \cdot \tan 35^{\circ } =12 \cdot \frac{AB}{12}$

$12 \tan 35^{\circ } =AB$ $\displaystyle 12 \tan 35^{\circ } =AB$

Now, set $AB = 12 \tan 35^{\circ }$ $\displaystyle AB = 12 \tan 35^{\circ }$ and AC=12 $\displaystyle AC=12$ in the area formula:

$a= \frac{1}{2}(AB)(AC)$ $\displaystyle a= \frac{1}{2}(AB)(AC)$

$= \frac{1}{2}(12 \tan 35^{\circ })\left (12\right )$ $\displaystyle = \frac{1}{2}(12 \tan 35^{\circ })\left (12\right )$

$=72 \tan 35^{\circ }$ $\displaystyle =72 \tan 35^{\circ }$,

the correct choice.

Flashcards: Tangent

Right

Refer to the triangle in the above diagram. Which of the following expressions correctly gives its area?

$32 \tan 35^{\circ}$ $\displaystyle 32 \tan 35^{\circ}$

None of the other choices gives the correct response.

$64 \tan 35^{\circ}$ $\displaystyle 64 \tan 35^{\circ}$

$\frac{64}{\tan 35^{\circ}}$ $\displaystyle \frac{64}{\tan 35^{\circ}}$

$\frac{32}{\tan 35^{\circ}}$ $\displaystyle \frac{32}{\tan 35^{\circ}}$

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Flashcards: Tangent

Refer to the triangle in the above diagram. Which of the following expressions correctly gives its area?

Your answer:

$32 \tan 35^{\circ}$ $\displaystyle 32 \tan 35^{\circ}$

Your answer:

None of the other choices gives the correct response.

Your answer:

$64 \tan 35^{\circ}$ $\displaystyle 64 \tan 35^{\circ}$

Your answer:

$\frac{64}{\tan 35^{\circ}}$ $\displaystyle \frac{64}{\tan 35^{\circ}}$

Correct answer:

$\frac{32}{\tan 35^{\circ}}$ $\displaystyle \frac{32}{\tan 35^{\circ}}$

Explanation:

$a= \frac{1}{2}(AB)(AC)$ $\displaystyle a= \frac{1}{2}(AB)(AC)$

We are given that AB = 8 $\displaystyle AB = 8$. $\overline{AB }$ $\displaystyle \overline{AB }$ is the leg opposite the angle $\angle C$ $\displaystyle \angle C$ and $\overline{AC}$ $\displaystyle \overline{AC}$ is its adjacent leg, we can find $\displaystyle AC$ using the tangent ratio:

$\tan (m \angle C) = \frac{AB}{AC}$ $\displaystyle \tan (m \angle C) = \frac{AB}{AC}$

Setting AB = 8 $\displaystyle AB = 8$ and $m \angle C = 35 ^{\circ }$ $\displaystyle m \angle C = 35 ^{\circ }$, we get

$\tan 35^{\circ } = \frac{8}{AC}$ $\displaystyle \tan 35^{\circ } = \frac{8}{AC}$

Solve for $\displaystyle AC$ by first, finding the reciprocal of both sides:

$\frac{1}{\tan 35^{\circ }} = \frac{AC}{8}$ $\displaystyle \frac{1}{\tan 35^{\circ }} = \frac{AC}{8}$

Now, multiply both sides by 8:

$\frac{1}{\tan 35^{\circ }} \cdot 8 = \frac{AC}{8} \cdot 8$ $\displaystyle \frac{1}{\tan 35^{\circ }} \cdot 8 = \frac{AC}{8} \cdot 8$

$\frac{8}{\tan 35^{\circ }} = AC$ $\displaystyle \frac{8}{\tan 35^{\circ }} = AC$

Now, set AB = 8 $\displaystyle AB = 8$ and $AC = \frac{8}{\tan 35^{\circ }}$ $\displaystyle AC = \frac{8}{\tan 35^{\circ }}$ in the area formula:

$a= \frac{1}{2}(AB)(AC)$ $\displaystyle a= \frac{1}{2}(AB)(AC)$

$= \frac{1}{2}(8)\left ( \frac{8}{\tan 35^{\circ}} \right )$ $\displaystyle = \frac{1}{2}(8)\left ( \frac{8}{\tan 35^{\circ}} \right )$

$= \frac{1}{2} \left (\frac{8}{1} \right )\left ( \frac{8}{\tan 35^{\circ}} \right )$ $\displaystyle = \frac{1}{2} \left (\frac{8}{1} \right )\left ( \frac{8}{\tan 35^{\circ}} \right )$

$= \frac{32}{\tan 35^{\circ}}$ $\displaystyle = \frac{32}{\tan 35^{\circ}}$,

the correct choice.

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