Question

Which of the following traingles is Cos B = 0.8?

Answer 1

Step-by-step explanation:

Given:

To find:

Which triangle is cos B = 0.8

For the 1st Triangle:

We have to first determine the value of side BC using the Pythagorean theorem as seen below;

`\begin{gathered} 5^2=4^2+BC^2 \\ 25=16+BC^2 \\ BC=√(25-16) \\ BC=√(9) \\ BC=3 \end{gathered}`

So cosine B will be;

`\begin{gathered} \cos B=\frac{adjacent\text{ }side\text{ }to\text{ }angle\text{ B}}{hypotenuse} \\ \cos B=(3)/(5) \\ \cos B=0.6 \end{gathered}`

For the 2nd Triangle:

`\begin{gathered} \cos B=\frac{adjacent\text{ }side\text{ }to\text{ }angle\text{ B}}{hypotenuse} \\ \cos B=(5)/(8) \\ \cos B=0.625 \end{gathered}`

For the 3rd Triangle:

We have to first determine the value of side AB using the Pythagorean theorem as seen below;

`\begin{gathered} AB^2=4^2+5^2 \\ AB^2=16+25 \\ AB=√(41) \end{gathered}`

So cosine B will be;

`\begin{gathered} \cos B=\frac{adjacent\text{ }side\text{ }to\text{ }angle\text{ B}}{hypotenuse} \\ \cos B=(4)/(√(41)) \\ \cos B=0.62 \end{gathered}`

For the 4th Triangle:

`\begin{gathered} \cos B=\frac{adjacent\text{ }side\text{ }to\text{ }angle\text{ B}}{hypotenuse} \\ \cos B=(4)/(5) \\ \cos B=0.8 \end{gathered}`

So in the below triangle cos B = 0.8