Respuesta :
Ok so cos(x) = sin(90-x) so using that you can get sin(7x-15) = sin(90-(3x+5)) so 7x-15 = -3x+85. 10x = 100 and x = 10. Use that to find the 2 angles are 55 and 35
we know that
in a right triangle
(∝ + β)=90°-------> by complementary angles
so
sin ∝ = cos β
[tex]sin(7x - 15) = cos(3x + 5)[/tex]
Let
∝=7x-15
β=3x+5
therefore
[tex](7x-15)+(3x+5)=90 \\ 10x-10=90 \\ 10x=90+10 \\ x= \frac{100}{10} [/tex]
[tex]x=10[/tex]°
∝=7x-15-----------> ∝=[tex]7*10-15[/tex]-------> ∝=55°
β=3x+5------------> β=[tex]3*10+5[/tex]-------> β=35°
the answer is
β=35°
in a right triangle
(∝ + β)=90°-------> by complementary angles
so
sin ∝ = cos β
[tex]sin(7x - 15) = cos(3x + 5)[/tex]
Let
∝=7x-15
β=3x+5
therefore
[tex](7x-15)+(3x+5)=90 \\ 10x-10=90 \\ 10x=90+10 \\ x= \frac{100}{10} [/tex]
[tex]x=10[/tex]°
∝=7x-15-----------> ∝=[tex]7*10-15[/tex]-------> ∝=55°
β=3x+5------------> β=[tex]3*10+5[/tex]-------> β=35°
the answer is
β=35°