Here, triangle AEN and ALG are similar.
We know that the ratio of corresponding sides of two similar triangles are equal. So,

Answer:
Step-by-step explanation:

<h2 /><h2>
<u>Consider</u></h2>

<h2>
<u>W</u><u>e</u><u> </u><u>K</u><u>n</u><u>o</u><u>w</u><u>,</u></h2>




So, on substituting all these values, we get




<h2>Hence,</h2>

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<h2>ADDITIONAL INFORMATION :-</h2>
Sign of Trigonometric ratios in Quadrants
- sin (90°-θ) = cos θ
- cos (90°-θ) = sin θ
- tan (90°-θ) = cot θ
- csc (90°-θ) = sec θ
- sec (90°-θ) = csc θ
- cot (90°-θ) = tan θ
- sin (90°+θ) = cos θ
- cos (90°+θ) = -sin θ
- tan (90°+θ) = -cot θ
- csc (90°+θ) = sec θ
- sec (90°+θ) = -csc θ
- cot (90°+θ) = -tan θ
- sin (180°-θ) = sin θ
- cos (180°-θ) = -cos θ
- tan (180°-θ) = -tan θ
- csc (180°-θ) = csc θ
- sec (180°-θ) = -sec θ
- cot (180°-θ) = -cot θ
- sin (180°+θ) = -sin θ
- cos (180°+θ) = -cos θ
- tan (180°+θ) = tan θ
- csc (180°+θ) = -csc θ
- sec (180°+θ) = -sec θ
- cot (180°+θ) = cot θ
- sin (270°-θ) = -cos θ
- cos (270°-θ) = -sin θ
- tan (270°-θ) = cot θ
- csc (270°-θ) = -sec θ
- sec (270°-θ) = -csc θ
- cot (270°-θ) = tan θ
- sin (270°+θ) = -cos θ
- cos (270°+θ) = sin θ
- tan (270°+θ) = -cot θ
- csc (270°+θ) = -sec θ
- sec (270°+θ) = cos θ
- cot (270°+θ) = -tan θ
Answer:
5.0 times 10 Superscript 6
Step-by-step explanation:
What is the value of StartFraction 1.6 times 10 Superscript 14 Baseline Over 3.2 times 10 Superscript 7 Baseline?
This is represented mathematically as:
1.6 × 10¹⁴/ 3.2 × 10⁷
Solving for this
1.6 × 10¹⁴/ 3.2 × 10⁷
= [1.6/3.2] × 10¹⁴-⁷
= 5 × 10⁶
= 5.0 times 10 Superscript 6
They will each get 1/6 of the pizza.
Answer:
Step-by-step explanation:
The ratio of its areas is equal to
Step-by-step explanation:
we know that
If two figures are similar, then the ratio of its areas is equal to the scale factor squared
In this problem the scale factor is equal to the ratio 10:3
Let
z-------> the scale factor
so
z2=(10/3)2=100/9