Math / Geometry

QuestionThe figure below contains two right triangles with one common side. The length of the segment $B C$ is $|B C|=2 \mathrm{~cm}$ , the length of the segment $A D$ is $|A D|=8 \mathrm{~cm}$ , and the angle $\angle D A B$ is $\angle D A B=53^{\circ}$ .
Find each of the following (note that the angles you plug into trigonometric functions must be in radians): The length of the line segment $B D:|B D|=6.389$ $\square$ cm . The tangent of angle $\theta=\angle B D C: \tan (\theta)=0.313$ The length of the line segement $C D:|C D|=1$ (1) cm .

Studdy Solution

STEP 1

What is this asking? We've got two right triangles snuggled together, sharing a side, and we need to find some lengths and an angle using what we know about trigonometry! Watch out! Remember to convert that angle from degrees to radians before using it in trig functions!
Also, make sure you're using the right trig function for the side you're looking for.

STEP 2

1. Find AB
2. Find BD
3. Find tan(θ)
4. Find CD

STEP 3

We're given the length of AD ( $|AD| = 8$ cm) and the angle $\angle DAB = 53^\circ$ .
We want to find the length of AB.
Since we have a right triangle and we know the angle and the side opposite the angle, and we want the side adjacent to the angle, we should use the tangent function!

STEP 4

First, let's convert our angle to radians.
Remember, $180^\circ = \pi$ radians.
So, $53^\circ$ is equal to \(53^\circ \cdot \frac{\pi}{180^\circ} \approx 0.925 radians.

STEP 5

Now, we know that $\tan(\angle DAB) = \frac{|AB|}{|AD|}$ .
Plugging in our values, we get $\tan(0.925) = \frac{|AB|}{8}$ .

STEP 6

To isolate $|AB|$ , we multiply both sides by 8: $8 \cdot \tan(0.925) = |AB|$ .
Calculating this gives us $|AB| \approx 8 \cdot 1.327 = 10.616$ cm.

STEP 7

Now we want to find the length of BD.
We have a right triangle with sides AB and BC, and BD is the hypotenuse.
Sounds like a job for the Pythagorean theorem!

STEP 8

The Pythagorean theorem states that $|BD|^2 = |AB|^2 + |BC|^2$ .
We know $|AB| \approx 10.616$ cm and $|BC| = 2$ cm.

STEP 9

Plugging in our values, we get $|BD|^2 \approx (10.616)^2 + (2)^2 \approx 112.699 + 4 = 116.699$ .

STEP 10

Taking the square root of both sides gives us $|BD| \approx \sqrt{116.699} \approx 10.798$ cm.

STEP 11

We're looking for $\tan(\theta)$ , where $\theta = \angle BDC$ .
In triangle BCD, $\tan(\theta) = \frac{|BC|}{|CD|}$ .
We know $|BC| = 2$ cm, but we need to find $|CD|$ .

STEP 12

We can use the Pythagorean theorem again with triangle BCD.
We know $|BC| = 2$ cm and we just found $|BD| \approx 10.798$ cm.
So, $|BD|^2 = |BC|^2 + |CD|^2$ .

STEP 13

Plugging in our values, we have $(10.798)^2 \approx (2)^2 + |CD|^2$ , which simplifies to $116.699 \approx 4 + |CD|^2$ .

STEP 14

Subtracting 4 from both sides gives us $112.596 \approx |CD|^2$ .
Taking the square root of both sides, we find $|CD| \approx \sqrt{112.596} \approx 10.611$ cm.

STEP 15

Now we can find $\tan(\theta)$ . $\tan(\theta) = \frac{|BC|}{|CD|} \approx \frac{2}{10.611} \approx 0.188$ .

STEP 16

$|BD| \approx 10.798$ cm $\tan(\theta) \approx 0.188$ $|CD| \approx 10.611$ cm

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Studdy Solution

STEP 1

STEP 2

1. Find AB
2. Find BD
3. Find tan(θ)
4. Find CD

STEP 3

We're given the length of AD ( $|AD| = 8$ cm) and the angle $\angle DAB = 53^\circ$ .
We want to find the length of AB.
Since we have a right triangle and we know the angle and the side opposite the angle, and we want the side adjacent to the angle, we should use the tangent function!

STEP 4

First, let's convert our angle to radians.
Remember, $180^\circ = \pi$ radians.
So, $53^\circ$ is equal to \(53^\circ \cdot \frac{\pi}{180^\circ} \approx 0.925 radians.

STEP 5

Now, we know that $\tan(\angle DAB) = \frac{|AB|}{|AD|}$ .
Plugging in our values, we get $\tan(0.925) = \frac{|AB|}{8}$ .

STEP 6

To isolate $|AB|$ , we multiply both sides by 8: $8 \cdot \tan(0.925) = |AB|$ .
Calculating this gives us $|AB| \approx 8 \cdot 1.327 = 10.616$ cm.

STEP 7

Now we want to find the length of BD.
We have a right triangle with sides AB and BC, and BD is the hypotenuse.
Sounds like a job for the Pythagorean theorem!

STEP 8

The Pythagorean theorem states that $|BD|^2 = |AB|^2 + |BC|^2$ .
We know $|AB| \approx 10.616$ cm and $|BC| = 2$ cm.

STEP 9

Plugging in our values, we get $|BD|^2 \approx (10.616)^2 + (2)^2 \approx 112.699 + 4 = 116.699$ .

STEP 10

Taking the square root of both sides gives us $|BD| \approx \sqrt{116.699} \approx 10.798$ cm.

STEP 11

We're looking for $\tan(\theta)$ , where $\theta = \angle BDC$ .
In triangle BCD, $\tan(\theta) = \frac{|BC|}{|CD|}$ .
We know $|BC| = 2$ cm, but we need to find $|CD|$ .

STEP 12

We can use the Pythagorean theorem again with triangle BCD.
We know $|BC| = 2$ cm and we just found $|BD| \approx 10.798$ cm.
So, $|BD|^2 = |BC|^2 + |CD|^2$ .

STEP 13

Plugging in our values, we have $(10.798)^2 \approx (2)^2 + |CD|^2$ , which simplifies to $116.699 \approx 4 + |CD|^2$ .

STEP 14

Subtracting 4 from both sides gives us $112.596 \approx |CD|^2$ .
Taking the square root of both sides, we find $|CD| \approx \sqrt{112.596} \approx 10.611$ cm.

STEP 15

Now we can find $\tan(\theta)$ . $\tan(\theta) = \frac{|BC|}{|CD|} \approx \frac{2}{10.611} \approx 0.188$ .

STEP 16

$|BD| \approx 10.798$ cm $\tan(\theta) \approx 0.188$ $|CD| \approx 10.611$ cm

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Studdy Solution

STEP 1

STEP 2

1. Find AB2. Find BD3. Find tan(θ)4. Find CD

STEP 3

STEP 4

First, let's **convert** our angle to radians.Remember, 180∘=π180^\circ = \pi180∘=π radians.So, 53∘53^\circ53∘ is equal to \(53^\circ \cdot \frac{\pi}{180^\circ} \approx 0.925 radians.

STEP 5

Now, we know that tan⁡(∠DAB)=∣AB∣∣AD∣\tan(\angle DAB) = \frac{|AB|}{|AD|}tan(∠DAB)=∣AD∣∣AB∣​.Plugging in our values, we get tan⁡(0.925)=∣AB∣8\tan(0.925) = \frac{|AB|}{8}tan(0.925)=8∣AB∣​.

STEP 6

To **isolate** ∣AB∣|AB|∣AB∣, we multiply both sides by **8**: 8⋅tan⁡(0.925)=∣AB∣8 \cdot \tan(0.925) = |AB|8⋅tan(0.925)=∣AB∣.Calculating this gives us ∣AB∣≈8⋅1.327=10.616|AB| \approx 8 \cdot 1.327 = 10.616∣AB∣≈8⋅1.327=10.616 cm.

STEP 7

Now we want to find the length of BD.We have a right triangle with sides AB and BC, and BD is the hypotenuse.Sounds like a job for the Pythagorean theorem!

STEP 8

The Pythagorean theorem states that ∣BD∣2=∣AB∣2+∣BC∣2|BD|^2 = |AB|^2 + |BC|^2∣BD∣2=∣AB∣2+∣BC∣2.We know ∣AB∣≈10.616|AB| \approx 10.616∣AB∣≈10.616 cm and ∣BC∣=2|BC| = 2∣BC∣=2 cm.

STEP 9

Plugging in our values, we get ∣BD∣2≈(10.616)2+(2)2≈112.699+4=116.699|BD|^2 \approx (10.616)^2 + (2)^2 \approx 112.699 + 4 = 116.699∣BD∣2≈(10.616)2+(2)2≈112.699+4=116.699.

STEP 10

Taking the square root of both sides gives us ∣BD∣≈116.699≈10.798|BD| \approx \sqrt{116.699} \approx 10.798∣BD∣≈116.699​≈10.798 cm.

STEP 11

We're looking for tan⁡(θ)\tan(\theta)tan(θ), where θ=∠BDC\theta = \angle BDCθ=∠BDC.In triangle BCD, tan⁡(θ)=∣BC∣∣CD∣\tan(\theta) = \frac{|BC|}{|CD|}tan(θ)=∣CD∣∣BC∣​.We know ∣BC∣=2|BC| = 2∣BC∣=2 cm, but we need to find ∣CD∣|CD|∣CD∣.

STEP 12

We can use the Pythagorean theorem again with triangle BCD.We know ∣BC∣=2|BC| = 2∣BC∣=2 cm and we just found ∣BD∣≈10.798|BD| \approx 10.798∣BD∣≈10.798 cm.So, ∣BD∣2=∣BC∣2+∣CD∣2|BD|^2 = |BC|^2 + |CD|^2∣BD∣2=∣BC∣2+∣CD∣2.

STEP 13

Plugging in our values, we have (10.798)2≈(2)2+∣CD∣2(10.798)^2 \approx (2)^2 + |CD|^2(10.798)2≈(2)2+∣CD∣2, which simplifies to 116.699≈4+∣CD∣2116.699 \approx 4 + |CD|^2116.699≈4+∣CD∣2.

STEP 14

Subtracting **4** from both sides gives us 112.596≈∣CD∣2112.596 \approx |CD|^2112.596≈∣CD∣2.Taking the square root of both sides, we find ∣CD∣≈112.596≈10.611|CD| \approx \sqrt{112.596} \approx 10.611∣CD∣≈112.596​≈10.611 cm.

STEP 15

Now we can find tan⁡(θ)\tan(\theta)tan(θ). tan⁡(θ)=∣BC∣∣CD∣≈210.611≈0.188\tan(\theta) = \frac{|BC|}{|CD|} \approx \frac{2}{10.611} \approx 0.188tan(θ)=∣CD∣∣BC∣​≈10.6112​≈0.188.

STEP 16

∣BD∣≈10.798|BD| \approx 10.798∣BD∣≈10.798 cm tan⁡(θ)≈0.188\tan(\theta) \approx 0.188tan(θ)≈0.188 ∣CD∣≈10.611|CD| \approx 10.611∣CD∣≈10.611 cm

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1. Find AB
2. Find BD
3. Find tan(θ)
4. Find CD

First, let's convert our angle to radians.
Remember, $180^\circ = \pi$ radians.
So, $53^\circ$ is equal to \(53^\circ \cdot \frac{\pi}{180^\circ} \approx 0.925 radians.

Now, we know that $\tan(\angle DAB) = \frac{|AB|}{|AD|}$ .
Plugging in our values, we get $\tan(0.925) = \frac{|AB|}{8}$ .

To isolate $|AB|$ , we multiply both sides by 8: $8 \cdot \tan(0.925) = |AB|$ .
Calculating this gives us $|AB| \approx 8 \cdot 1.327 = 10.616$ cm.

Now we want to find the length of BD.
We have a right triangle with sides AB and BC, and BD is the hypotenuse.
Sounds like a job for the Pythagorean theorem!

The Pythagorean theorem states that $|BD|^2 = |AB|^2 + |BC|^2$ .
We know $|AB| \approx 10.616$ cm and $|BC| = 2$ cm.

Plugging in our values, we get $|BD|^2 \approx (10.616)^2 + (2)^2 \approx 112.699 + 4 = 116.699$ .

Taking the square root of both sides gives us $|BD| \approx \sqrt{116.699} \approx 10.798$ cm.

We're looking for $\tan(\theta)$ , where $\theta = \angle BDC$ .
In triangle BCD, $\tan(\theta) = \frac{|BC|}{|CD|}$ .
We know $|BC| = 2$ cm, but we need to find $|CD|$ .

We can use the Pythagorean theorem again with triangle BCD.
We know $|BC| = 2$ cm and we just found $|BD| \approx 10.798$ cm.
So, $|BD|^2 = |BC|^2 + |CD|^2$ .

Plugging in our values, we have $(10.798)^2 \approx (2)^2 + |CD|^2$ , which simplifies to $116.699 \approx 4 + |CD|^2$ .

Subtracting 4 from both sides gives us $112.596 \approx |CD|^2$ .
Taking the square root of both sides, we find $|CD| \approx \sqrt{112.596} \approx 10.611$ cm.

Now we can find $\tan(\theta)$ . $\tan(\theta) = \frac{|BC|}{|CD|} \approx \frac{2}{10.611} \approx 0.188$ .

$|BD| \approx 10.798$ cm $\tan(\theta) \approx 0.188$ $|CD| \approx 10.611$ cm