Find the valu of x and y

Mathematics

2 answers:

irinina [24]3 years ago

5 0

Answer:

x=63; y=27

Step-by-step explanation:

The straight line of the 117 is 180 and so you subtract 180 to 117 to get 63 which is x and then subtract 90 and 63 to get 27.

iren [92.7K]3 years ago

3 0

Answer:
x= 63
y= 27

Working out:
x : 180-117=63
y : 90-63=27

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Four consecutive even integers are such that the sum of the second and fourth is 76. What is the sum of the first and third inte

Arada [10]

Answer:

Step-by-step explanation:

Consecutive even integers have a difference of 2 between them

let n be the first integer, then

n + 2, n + 4, n + 6 are the second, third and fourth, then

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3 years ago

How is this solved using trig identities (sum/difference)?

GenaCL600 [577]

FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

$cos \beta = -\frac{1}{2} \sqrt{3}$

$tan \beta= \frac{1}{3} \sqrt{3}$

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
$sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )$
$sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )$
$sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})$

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
$cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )$
$cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )$
$cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )$

Find tan (α - β)
$tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}$

Simplify the denominator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the numerator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$
$tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the fraction
$tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})$
$tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}$