Math / Trigonometry

QuestionLet $\theta$ be an angle in quadrant I such that $\sin \theta=\frac{7}{10}$ . Find the exact values of $\sec \theta$ and $\tan \theta$ .

Studdy Solution

STEP 1

What is this asking? If the sine of an angle $\theta$ in the first quadrant is $7/10$ , what are the secant and tangent of that angle? Watch out! Remember that trigonometric functions relate ratios of sides in a right-angled triangle, and their values depend on which quadrant the angle is in.

STEP 2

1. Visualize with a Triangle
2. Calculate the Adjacent Side
3. Find Secant and Tangent

STEP 3

Let's draw a right-angled triangle in the first quadrant where $\theta$ is one of the acute angles.
Since $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$ , and we're given $\sin \theta = \frac{7}{10}$ , we can label the opposite side as $7$ and the hypotenuse as $10$ .

STEP 4

This triangle helps us visualize the problem and makes it easier to find the other trigonometric ratios.
Remember SOH CAH TOA!

STEP 5

We can find the length of the adjacent side using the Pythagorean theorem: $a^2 + b^2 = c^2$ , where $a$ and $b$ are the legs and $c$ is the hypotenuse.
In our case, the opposite side is $7$ , the hypotenuse is $10$ , and we want to find the adjacent side, let's call it $x$ .

STEP 6

So, we have $7^2 + x^2 = 10^2$ , which simplifies to $49 + x^2 = 100$ .
Subtracting 49 from both sides gives us $x^2 = 51$ .
Taking the positive square root (since lengths are positive) gives us $x = \sqrt{51}$ .
So, the adjacent side has length $\sqrt{51}$ .

STEP 7

Now that we know all three sides of the triangle, we can find $\sec \theta$ and $\tan \theta$ .
Remember that $\sec \theta = \frac{1}{\cos \theta}$ and $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ .

STEP 8

Therefore, $\cos \theta = \frac{\sqrt{51}}{10}$ , and $\sec \theta = \frac{10}{\sqrt{51}}$ .
We can rationalize the denominator by multiplying the numerator and denominator by $\sqrt{51}$ , which gives us $\sec \theta = \frac{10\sqrt{51}}{51}$ .

STEP 9

Next, recall that $\tan \theta = \frac{\text{opposite}}{\text{adjacent}}$ .
In our triangle, the opposite side is $7$ and the adjacent side is $\sqrt{51}$ , so $\tan \theta = \frac{7}{\sqrt{51}}$ . Rationalizing the denominator gives us $\tan \theta = \frac{7\sqrt{51}}{51}$ .

STEP 10

We found that $\sec \theta = \frac{10\sqrt{51}}{51}$ and $\tan \theta = \frac{7\sqrt{51}}{51}$ .

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