sin2x =12/13
cos2x = 5/13
tan2x = 12/5
STEP - BY - STEP EXPLANATION
What to find?
• sin2x
,
• cos2x
,
• tan2x
Given:
tanx = 2/3 = opposite / adjacent
We need to first make a sketch of the given problem.
Let h be the hypotenuse.
We need to find sinx and cos x, but to find sinx and cosx, first determine the value of h.
Using the Pythagoras theorem;
hypotenuse² = opposite² + adjacent²
h² = 2² + 3²
h² = 4 + 9
h² =13
Take the square root of both-side of the equation.
h =√13
This implies that hypotenuse = √13
We can now proceed to find the values of ainx and cosx.
Using the trigonometric ratio;
![\sin x=\frac{opposite}{\text{hypotenuse}}=\frac{2}{\sqrt[]{13}}](https://tex.z-dn.net/?f=%5Csin%20x%3D%5Cfrac%7Bopposite%7D%7B%5Ctext%7Bhypotenuse%7D%7D%3D%5Cfrac%7B2%7D%7B%5Csqrt%5B%5D%7B13%7D%7D)
![\cos x=\frac{adjacent}{\text{hypotenuse}}=\frac{3}{\sqrt[]{13}}](https://tex.z-dn.net/?f=%5Ccos%20x%3D%5Cfrac%7Badjacent%7D%7B%5Ctext%7Bhypotenuse%7D%7D%3D%5Cfrac%7B3%7D%7B%5Csqrt%5B%5D%7B13%7D%7D)
And we know that tanx =2/3
From the trigonometric identity;
sin 2x = 2sinxcosx
Substitute the value of sinx , cosx and then simplify.
![\sin 2x=2(\frac{2}{\sqrt[]{13}})(\frac{3}{\sqrt[]{13}})](https://tex.z-dn.net/?f=%5Csin%202x%3D2%28%5Cfrac%7B2%7D%7B%5Csqrt%5B%5D%7B13%7D%7D%29%28%5Cfrac%7B3%7D%7B%5Csqrt%5B%5D%7B13%7D%7D%29)

Hence, sin2x = 12/13
cos2x = cos²x - sin²x
Substitute the value of cosx, sinx and simplify.
![\begin{gathered} \cos 2x=(\frac{3}{\sqrt[]{13}})^2-(\frac{2}{\sqrt[]{13}})^2 \\ \\ =\frac{9}{13}-\frac{4}{13} \\ =\frac{5}{13} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ccos%202x%3D%28%5Cfrac%7B3%7D%7B%5Csqrt%5B%5D%7B13%7D%7D%29%5E2-%28%5Cfrac%7B2%7D%7B%5Csqrt%5B%5D%7B13%7D%7D%29%5E2%20%5C%5C%20%20%5C%5C%20%3D%5Cfrac%7B9%7D%7B13%7D-%5Cfrac%7B4%7D%7B13%7D%20%5C%5C%20%3D%5Cfrac%7B5%7D%7B13%7D%20%5Cend%7Bgathered%7D)
Hence, cos2x = 5/13
tan2x = 2tanx / 1- tan²x






OR

Hence, tan2x = 12/5
Therefore,
sin2x =12/13
cos2x = 5/13
tan2x = 12/5
Total distance 5 km; at 5km / 0.65 h =
Second part distance: x; at 6 km/h, during t2
First part distance: 5 - x; at 8.75 km/h, during t1
V = d/t ⇒ t = d/V
t2 = x/6
t1=[5-x]/8.75
t2 + t1 = 0.65
x/6 + [5-x]/8.75 = 0.65
x/6 + 5/8.75 - x/8.75 = 13/20
x/6 - x/8.75 = 13/20 - 5/8.75
x/6 - 4x/35 =13/20 - 20/35
35x - 24x = (35*6)(35*13 - 20*20)/(20*35)
11 x = 16.5
x = 16.5/11 = 1.5 km
Answer: 2/6 or 1/3
Step-by-step explanation:
SCHOOL has 6 letters in it and two O's. Therefore, it would be 2/6 or simplified it would be 1/3.
That was FUN!
Complementary angles = 90
x+y =90
y=x+24
substitute this in to the equation
x+x+24=90
2x+24=90
subtract 24 from each side
2x = 66
divide by 2
x=33
y = 33+24
y=57
Answer: 33, 57
Answer:
263
Step-by-step explanation:
take the start amount 315 and subtract 78 then add 26 which would amount to 263