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<u>Note:</u>
The answer in the given link is answered by me.
Answer:
(a) 4
(b) 2√3
(c) 60°
(d) 120°
Step-by-step explanation:
(a) The relationship between tangents and secants is ...
CB^2 = CD·CA
Filling in the given values, we find ...
CB^2 = 2·(2+6) = 16
CB = √16 = 4
The length of BC is 4 units.
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(b) Triangle ABC is a right triangle, so the sides of it satisfy the Pythagorean theorem.
CA^2 = CB^2 +AB^2
8^2 = 16 +AB^2
AB = √48 = 4√3
The radius is half the length of AB, so the radius is 2√3.
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(c) The measure of angle C can be determined from the cosine relation:
cos(C) = CB/CA = 4/8 = 1/2
C = arccos(1/2) = 60°
The measure of angle C is 60°.
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(d) Arc AD is intercepted by angle ABD, which has the same measure as angle C. Hence the measure of arc AD is twice the measure of angle C.
The measure of arc AD is 120°.
The value of the angle θ required in the water main for the turn from north to east is 87.40°.
According to the statement
We have to find that the value of the angle.
So, For this purpose, we know that the
An angle is created when two straight lines or rays meet at a standard endpoint.
From the given information:
a main is to be constructed with a 20% grade within the north direction and a tenth grade within the east direction.
Then
Let n = North
and e = east
Then
using vectors,
n = j + 0.20 k
e = i + 0.25 k
And the
n · e = 1x0 + 0x1 + 0.20×0.25 = 0.05
Now take magnitude then
|n|·|e| = i(1² + 0.20²)·i(1²+0.25²) = 1.105
Now,
E° = arccos((n · e)/|n|·|e|)
E° = arccos(0.05/1.105)
E° = arccos (0.0452)
E° = 87.40°
So, The value of the angle θ required in the water main for the turn from north to east is 87.40°.
Learn more about angle here
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