Question

1) iven sinx=45 and cosx=35 .

Answer 1

Answer:

Part 1) $tan(x)=\frac{4}{3}$

Part 2) $sin(A)=(8/17)$

Part 3) $sin(x)=0.87$

Part 4) $x=25\°$

Step-by-step explanation:

Part 1) Note In this problem sinx should be 4/5 not 45 and cosx should be 3/5 not 35

Given sinx=4/5 and cosx=3/5 (see the note)

What is ratio for ​ tanx ​ ?

we know that

$tan(x)=\frac{sin(x)}{cos(x)}$

substitute the values

$tan(x)=\frac{(4/5)}{(3/5)}$

$tan(x)=\frac{4}{3}$

Part 2) ∠A is an acute angle in a right triangle

Note In this problem cosA should be 15/17 not 1517

Given that cosA=15/17, what is the ratio for sinA?

we know that

$sin^{2}(A)+cos^{2}(A)=1$

substitute the value of cos(A) and solve for sin(A)

$sin^{2}(A)+(15/17)^{2}=1$

$sin^{2}(A)=1-(225/289)$

$sin^{2}(A)=(64/289)$

$sin(A)=(8/17)$

Part 3) Given sinx=0.5 , what is cosx ?

we know that

$sin^{2}(x)+cos^{2}(x)=1$

substitute the value of sin(x) and solve for cos(x)

$sin^{2}(x)+(0.5)^{2}=1$

$sin^{2}(x)=1-0.25$

$sin^{2}(x)=0.75$

$sin(x)=0.87$

Part 4) What is the value of x?

sin(x+22)°=cos(2x−7)°

we know that

if $sin(A)=cos(B)$

then

$A+B=90\°$ -----> by complementary angles

so

in this problem

$(x+22)+(2x-7)=90\°$

solve for x

$3x+15\°=90\°$

$3x=90\°-15\°$

$x=75\°/3=25\°$