The trigonometric function gives the ratio of different sides of a right-angle triangle. The given problems can be solved as given below.
<h3>What are Trigonometric functions?</h3>
The trigonometric function gives the ratio of different sides of a right-angle triangle.

where perpendicular is the side of the triangle which is opposite to the angle, and the hypotenuse is the longest side of the triangle which is opposite to the 90° angle.
1st.) x = 5 /Sin(30°)
x = 10
!) sin(45°) = 4/x
x = 4/sin(45°)
x = 4√2
I) Cos(45°) = √3 / x
x = √3 / Cos(45°)
x = √6
E) Tan(60°) = 3√3 / x
x = 3√3 / 3
W) For isosceles right-triangle, the angle made by the legs and the hypotenuse is always 45°.
x = 45°
N) x² + x² = (7√2)²
x = 7
V) Tan(60°) = 7 / x
x = 7√3/3
K) x² + x² = (9)²
x = 9/√2
Y) Sin(60°) = 7√3/x
x = 14
M) Sin(30°) = x/11
x = 11/2
T) Sin(45°) = x/√10
x = √5
A) x + 2x + 90° = 180°
x = 30°
O) Sin(45°) = √2 / x
x = 2
R) Tan(30°) = x / 4
x = 4/√3 = 4√3 / 3
S) Sin(60°) = x / (10/3)
x = 5√3 / 3
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Answer:
rational
Step-by-step explanation:
First, find the gradient/slope:
Use slope formula:
m=(y2-y1)/(x2-x1)
=(11-5)/(3-1)
=6/2
=3
Then use the line equation formula:
y=mx+c
You can substitute (1,5) if you like, also must substitute the slope as well!
5=3x1+c
c=2
Then find the full equation, which gives you the answer:
y=3x+2
4 cos² x - 3 = 0
4 cos² x = 3
cos² x = 3/4
cos x = ±(√3)/2
Fixing the squared cosine doesn't discriminate among quadrants. There's one in every quadrant
cos x = ± cos(π/6)
Let's do plus first. In general, cos x = cos a has solutions x = ±a + 2πk integer k
cos x = cos(π/6)
x = ±π/6 + 2πk
Minus next.
cos x = -cos(π/6)
cos x = cos(π - π/6)
cos x = cos(5π/6)
x = ±5π/6 + 2πk
We'll write all our solutions as
x = { -5π/6, -π/6, π/6, 5π/6 } + 2πk integer k
5 and 6 are true but 7 is not true. If you divide an odd number by 2 it's not going to be half of it