Question

11.
Use ∆ABC to find the value of cos B.
21.
Use ∆ABC to find the value of tan B.

(Shorter one is number 11)

Answer 1

The first thing we are going to do for this case is to define the cosine and the tangent:
 For the cosine we have:
 $Cos (x) = \frac{C.A}{h}$
 Where,
 C.A: adjacent leg
 h: hypotenuse
 On the other hand we have that the tangent is given by:
 Where,
 $Tan (x) = \frac{C.O}{C.A}$
 Where,
 C.O: opposite leg
 C.A: adjacent leg
 We have then:
 
 Part A
 $Cos (B) = \frac{15}{39}$
 Simplifying:
 $Cos (B) = \frac{5}{13}$

 Part B
 $Tan (B) = \frac{40}{9}$

Answer 2

Answer:

cos B = $\frac{5}{13}$

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

Step-by-step explanation:

• cos Θ = adjacent / hypotenuse
• sin Θ = opposite / hypotenuse
• tan Θ = opposite / adjacent

11) cos B

• the adjacent leg is 15
• the hypotenuse is 39

since cos Θ = adjacent / hypotenuse

then cos B = $\frac{15}{39}$

dividing both the numerator and denominator by 3 in order to simplify the answer

cos B = $\frac{5}{13}$

For the other picture

• the adjacent leg is 9
• the hypotenuse is 41

since cos Θ = adjacent / hypotenuse

then cos B = $\frac{9}{41}$

21) tan B

• the opposite leg is 40
• the adjacent leg is 9

since tan Θ = opposite / adjacent

then tan B = $\frac{40}{9}$

For the other picture

• the opposite leg is 36
• the adjacent leg is 15

since tan Θ = opposite / adjacent

then tan B = $\frac{36}{15}$

simplifying (diving numerator and denominator by 3)

tan B = $\frac{12}{5}$