<u>Answer
</u>
4
<u>Explanation
</u>
AP is a tangent meeting a secant BP at P outside the circle.
(PA)² = (PE)(PB)
PA = 6 and chord BE= 5.
Let PE = x
6² = x(x+5)
36 = x² + 5x
x² + 5x + 2.5² = 36 + 2.5²
(x + 2.5)² = 42.25
(x + 2.5) = √ 42.25 = 6.5
x = 6.5 - 2.5 = 4 or x = -6.5 - 2.5 = -9
But since length cannot be negative, x = 4.
You can consider the tree perpendicular to its shadow, therefore you get a triangle rectangle of which you know the two legs and you need to find the angle between the shadow and the hypothenuse, opposite to the tree.
In this case, the angle α can be easily found with trigonometry:
tan α = (opposite side / adjacent side)
Therefore:
α = tan⁻¹ (tree / shadow) = tan⁻¹ (10 / 14) = 35°
NOTE:
tan⁻¹ is sometimes found written as arctan.
For solving steps use photomath
Answer:
Try the first one I think and I remember when I did this one but I forgot the answer to this
Answer:
0.25844930417
Step-by-step explanation:
I hope this is the answer but I dont think it is.
