Question $$ 1.1 an heta=\frac{9}{40} $$ and $$ \cos heta

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Question $$ 1.1 	an 	heta=\frac{9}{40} $$ and $$ \cos 	heta<0 $$. Find the value of $$ \sin 	heta+\cos 	heta $$ by using a sketch and not finding the value of $$ 	heta $$.

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Given:
$$
\tan \theta = \frac{9}{40} \quad \text{and} \quad \cos \theta < 0
$$

**Step 1: Determine the Quadrant**

- $$\tan \theta = \frac{9}{40}$$ is positive, which means $$\theta$$ is either in the 1st or 3rd quadrant.
- $$\cos \theta < 0$$ indicates that $$\theta$$ is either in the 2nd or 3rd quadrant.

**Intersection:** $$\theta$$ must be in the **3rd quadrant**.

**Step 2: Construct the Right Triangle**

In the 3rd quadrant:
- **Opposite side (y):** Negative
- **Adjacent side (x):** Negative
- **Hypotenuse (r):** Positive (always)

Given $$\tan \theta = \frac{9}{40}$$:
$$
\tan \theta = \frac{|\text{opposite}|}{|\text{adjacent}|} = \frac{9}{40}
$$
Thus:
$$
\text{Opposite side} = -9, \quad \text{Adjacent side} = -40
$$
$$
\text{Hypotenuse} = \sqrt{(-9)^2 + (-40)^2} = \sqrt{81 + 1600} = \sqrt{1681} = 41
$$

**Step 3: Find $$\sin \theta$$ and $$\cos \theta$$**

$$
\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{-9}{41}
$$
$$
\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{-40}{41}
$$

**Step 4: Calculate $$\sin \theta + \cos \theta$$**

$$
\sin \theta + \cos \theta = \frac{-9}{41} + \frac{-40}{41} = \frac{-49}{41}
$$

**Final Answer:**
$$
\sin \theta + \cos \theta = -\frac{49}{41}
$$
$$ \sin \theta + \cos \theta = -\frac{49}{41} $$

Question and . Find the value of by using a sketch and
not finding the value of .

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Question and . Find the value of by using a sketch and not finding the value of .