Question

HELP IS FOR TODAY GIVE ME THE ANSWER AND HOW TO DO IT

Answer 1

Answer:

Part 1)

a) $sin(A)=\frac{9}{41}$

b) $sin(B)=\frac{40}{41}$

c) $cos(A)=\frac{40}{41}$

d) $cos(B)=\frac{9}{41}$

e) $tan(A)=\frac{9}{40}$

f) $tan(B)=\frac{40}{9}$

Part 2) $x=9.4\ units$ (see the explanation)

Part 3) $x=12.5\ units$ (see the explanation)

Part 4) $x=4.5\ units$ (see the explanation)

Part 5) $x=41.8^o$ (see the explanation)

Part 6) $x=50.2^o$ (see the explanation)

Part 7) $x=56.9^o$ (see the explanation)

Step-by-step explanation:

Part 1)

step 1

Find sin(A)

we know that

$sin(A)=\frac{BC}{AB}$ ----> by SOH (opposite side divided by the hypotenuse)

substitute the given values

$sin(A)=\frac{9}{41}$

step 2

Find sin(B)

we know that

$sin(B)=\frac{AC}{AB}$ ----> by SOH (opposite side divided by the hypotenuse)

substitute the given values

$sin(B)=\frac{40}{41}$

step 3

Find cos(A)

we know that

If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa

In this problem

Angle A and Angle B are complementary

so

$cos(A)=sin(B)$  ----> by complementary angles

therefore

$cos(A)=\frac{40}{41}$

step 4

Find cos(B)

we know that

If two angles are complementary, the cofunction identities state that the sine of one equals the cosine of the other and vice versa

In this problem

Angle A and Angle B are complementary

so

$cos(B)=sin(A)$  ----> by complementary angles

therefore

$cos(B)=\frac{9}{41}$

step 5

Find tan(A)

we know that

$tan(A)=\frac{BC}{AC}$ ----> by TOA (opposite side divided by adjacent side)

substitute the given values

$tan(A)=\frac{9}{40}$

step 6

Find tan(B)

we know that

$tan(B)=\frac{AC}{BC}$ ----> by TOA (opposite side divided by adjacent side)

substitute the given values

$tan(B)=\frac{40}{9}$

Part 2) we know that

$tan(58^o)=\frac{15}{x}$ ----> by TOA (opposite side divided by adjacent side)

solve for x

$x=\frac{15}{tan(58^o)}$

$x=9.4\ units$

Part 3) we know that

$sin(53^o)=\frac{10}{x}$ ----> by SOH (opposite side divided by the hypotenuse)

solve for x

$x=\frac{10}{sin(53^o)}$

$x=12.5\ units$

Part 4) we know that

$cos(41^o)=\frac{x}{6}$ ----> by CAH (adjacent side divided by the hypotenuse)

solve for x

$x=(6)cos(41^o)$

$x=4.5\ units$

Part 5) we know that

$sin(x)=\frac{4}{6}$ ----> by SOH (opposite side divided by the hypotenuse)

solve for angle x

$x=sin^{-1}(\frac{4}{6})$

$x=41.8^o$

Part 6) we know that

$tan(x)=\frac{30}{25}$ ----> by TOA (opposite side divided by adjacent side)

solve for angle x

$x=tan^{-1}(\frac{30}{25})$

$x=50.2^o$

Part 7) we know that

$cos(x)=\frac{12}{22}$ ----> by CAH (adjacent side divided by the hypotenuse)

solve for angle x

$x=cos^{-1}(\frac{12}{22})$

$x=56.9^o$