Using Exact trig Values Free Worksheet | Printable PDF Worksheets - Free Printable

Let's solve this step-by-step, starting with Section A, then moving on to Section B and Section C.

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✔ Section A: Complete the table using exact trig values

We are given two special right triangles:
- A 30°–60°–90° triangle with sides $1$, $\sqrt{3}$, and $2$
- A 45°–45°–90° triangle with sides $1$, $1$, and $\sqrt{2}$

We'll use these to fill in the exact trig values for $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, and $90^\circ$.

| Angle | $0^\circ$ | $30^\circ$ | $45^\circ$ | $60^\circ$ | $90^\circ$ |
|-------|-----------|------------|------------|------------|------------|
| sin | 0 | | | | 1 |
| cos | 1 | | | | 0 |
| tan | 0 | | | | undefined (or ∞) |

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#### 🔹 Using the 30°–60°–90° triangle:

Sides: opposite $30^\circ = 1$, opposite $60^\circ = \sqrt{3}$, hypotenuse = 2

- $\sin(30^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{2}$
- $\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2}$
- $\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$

For $60^\circ$:
- $\sin(60^\circ) = \frac{\sqrt{3}}{2}$
- $\cos(60^\circ) = \frac{1}{2}$
- $\tan(60^\circ) = \frac{\sqrt{3}}{1} = \sqrt{3}$

#### 🔹 Using the 45°–45°–90° triangle:

Sides: both legs = 1, hypotenuse = $\sqrt{2}$

- $\sin(45^\circ) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
- $\cos(45^\circ) = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
- $\tan(45^\circ) = \frac{1}{1} = 1$

#### 🔹 At $0^\circ$ and $90^\circ$:

- $\sin(0^\circ) = 0$, $\cos(0^\circ) = 1$, $\tan(0^\circ) = 0$
- $\sin(90^\circ) = 1$, $\cos(90^\circ) = 0$, $\tan(90^\circ)$ is undefined (since $\cos(90^\circ) = 0$)

Now complete the table:

| | $0^\circ$ | $30^\circ$ | $45^\circ$ | $60^\circ$ | $90^\circ$ |
|-------|-----------|------------|------------|------------|------------|
| sin | 0 | $\frac{1}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{3}}{2}$ | 1 |
| cos | 1 | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{1}{2}$ | 0 |
| tan | 0 | $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ | 1 | $\sqrt{3}$ | undefined |

✔ Section A Complete

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✔ Section B: Give the exact values

We’ll now use the exact values from Section A.

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1) $\sin 30 + \cos 60$

$$
= \frac{1}{2} + \frac{1}{2} = 1
$$

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2) $\tan 45 + \cos 60$

$$
= 1 + \frac{1}{2} = \frac{3}{2}
$$

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3) $4\sin 30 + \cos 0$

$$
= 4 \cdot \frac{1}{2} + 1 = 2 + 1 = 3
$$

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4) $2\sin 60$

$$
= 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}
$$

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5) $8\cos 45$

$$
= 8 \cdot \frac{\sqrt{2}}{2} = 4\sqrt{2}
$$

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6) $(\sin 45)^2 + 3\cos 60$

$$
= \left(\frac{\sqrt{2}}{2}\right)^2 + 3 \cdot \frac{1}{2} = \frac{2}{4} + \frac{3}{2} = \frac{1}{2} + \frac{3}{2} = 2
$$

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7) $(3\tan 30)^2 + (\tan 60)^2$

First:
- $\tan 30 = \frac{1}{\sqrt{3}}$, so $3\tan 30 = 3 \cdot \frac{1}{\sqrt{3}} = \sqrt{3}$
- $(\sqrt{3})^2 = 3$
- $\tan 60 = \sqrt{3}$, so $(\sqrt{3})^2 = 3$

So:
$$
3 + 3 = 6
$$

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8) $5\sin 60 - 3\cos 30$

$$
= 5 \cdot \frac{\sqrt{3}}{2} - 3 \cdot \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} - \frac{3\sqrt{3}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3}
$$

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9) $9\sin 30 - 3\sin 90 + 4(\cos 45)^2$

Break it down:
- $\sin 30 = \frac{1}{2}$ → $9 \cdot \frac{1}{2} = \frac{9}{2}$
- $\sin 90 = 1$ → $3 \cdot 1 = 3$
- $\cos 45 = \frac{\sqrt{2}}{2}$ → $(\cos 45)^2 = \frac{2}{4} = \frac{1}{2}$ → $4 \cdot \frac{1}{2} = 2$

Now:
$$
\frac{9}{2} - 3 + 2 = \frac{9}{2} - 1 = \frac{7}{2}
$$

✔ Section B Complete

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✔ Section C: Find missing angles using ratios

Use inverse trig functions: $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$

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1) Triangle: adjacent = 3 cm, hypotenuse = 6 cm, angle $a$

$$
\cos a = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{3}{6} = \frac{1}{2}
\Rightarrow a = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ
$$

✔ $a = 60^\circ$

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2) Triangle: opposite = 5 cm, hypotenuse = 10 cm, angle $b$

$$
\sin b = \frac{5}{10} = \frac{1}{2} \Rightarrow b = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ
$$

✔ $b = 30^\circ$

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3) Triangle: opposite = 2 cm, hypotenuse = $2\sqrt{2}$ cm, angle $c$

$$
\sin c = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \Rightarrow c = \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = 45^\circ
$$

✔ $c = 45^\circ$

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4) Triangle: opposite = $3\sqrt{3}$ cm, adjacent = 3 cm, angle $d$

$$
\tan d = \frac{3\sqrt{3}}{3} = \sqrt{3} \Rightarrow d = \tan^{-1}(\sqrt{3}) = 60^\circ
$$

✔ $d = 60^\circ$

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5) Triangle: opposite = $4\sqrt{2}$ cm, hypotenuse = 8 cm, angle $e$

$$
\sin e = \frac{4\sqrt{2}}{8} = \frac{\sqrt{2}}{2} \Rightarrow e = \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = 45^\circ
$$

✔ $e = 45^\circ$

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6) Triangle: opposite = $6\sqrt{3}$ cm, hypotenuse = 12 cm, angle $f$

$$
\sin f = \frac{6\sqrt{3}}{12} = \frac{\sqrt{3}}{2} \Rightarrow f = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = 60^\circ
$$

✔ $f = 60^\circ$

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✔ Final Answers Summary

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#### Section A Table:

| | $0^\circ$ | $30^\circ$ | $45^\circ$ | $60^\circ$ | $90^\circ$ |
|-------|-----------|------------|------------|------------|------------|
| sin | 0 | $\frac{1}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{3}}{2}$ | 1 |
| cos | 1 | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{1}{2}$ | 0 |
| tan | 0 | $\frac{\sqrt{3}}{3}$ | 1 | $\sqrt{3}$ | undefined |

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#### Section B: Exact Values

1) $1$
2) $\frac{3}{2}$
3) $3$
4) $\sqrt{3}$
5) $4\sqrt{2}$
6) $2$
7) $6$
8) $\sqrt{3}$
9) $\frac{7}{2}$

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#### Section C: Missing Angles

1) $a = 60^\circ$
2) $b = 30^\circ$
3) $c = 45^\circ$
4) $d = 60^\circ$
5) $e = 45^\circ$
6) $f = 60^\circ$

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