1. In right triangle ABC, C is the right angle. Given m2. In right triangle ABC, C is the right angle. Which of the following is cos B if sin A=0.4?

[tex]\Sin A=0.4=\frac{4}{10}=\frac{2}{5}\\\\In\ right\ triangle\\\\\sin A=\frac{Perpendicular}{Hypotenuse}=\frac{BC}{AB}=\frac{2}{5}\\\\Then\ \ \cos B=\frac{Base}{Hypotenuse}=\frac{BC}{AB}=\frac{2}{5}=0.4[/tex]

Answer Link

calculista calculista · Answer 2 · 2020-02-13T09:46:59+01:00

Answer:

Part a)

[tex]c=9.3\ units\\b=7.2\ units[/tex]

Part b) [tex]cos(B)=0.4[/tex] see the explanation

Step-by-step explanation:

The correct question is

In right triangle ABC, C is the right angle. Given measure of angle A = 40 degrees and a =6

Part a) which of the following are the lengths of the remaining two side, rounded to the nearest tenth?

Part b) Which of the following is cos B if sin A=0.4?

see the attached figure to better understand the problem

Part a)

step 1

Find the length of side c

Applying the law of sines

[tex]\frac{a}{sin(A)}=\frac{c}{sin(C)}[/tex]

we have

[tex]a=6\ units\\A=40^o\\C=90^o[/tex]

substitute

[tex]\frac{6}{sin(40^o)}=\frac{c}{sin(90^o)}[/tex]

solve for c

[tex]c=\frac{6}{sin(40^o)}=9.3\ units[/tex]

step 2

Find the length of side b

In the right triangle ABC

[tex]tan(40^o)=\frac{BC}{AC}[/tex] ----> by TOA (opposite side divided by the adjacent side)

substitute the values

[tex]tan(40^o)=\frac{6}{AC}[/tex]

[tex]AC=\frac{6}{tan(40^o)}=7.2\ units[/tex]

therefore

[tex]b=7.2\ units[/tex]

Part 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

therefore

the sine of angle A equals the cosine of angle B

we have

sin(A)=0.4

so

cos(B)=0.4