Question

In triangle abc, cos A= -0.6. Find A and Tan A

Answer 1

Answer:

Part 1) $A=126.87^o$

Part 2) $tan(A)=-\frac{4}{3}$

Step-by-step explanation:

we have

$cos(A)=-0.6$

The cos(A) is negative, that means that the angle A in the triangle ABC is an obtuse angle and the value of the sin(A) is positive

The angle A lie on the II Quadrant

step 1

Find the measure of angle A

$cos(A)=-0.6$

using a calculator

$A=cos^{-1}(-0.6)=126.87^o$

step 2

Find the sin(A)

we know that

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

substitute the value of cos(A)

$sin^2(A)+(-0.6)^2=1$

$sin^2(A)=1-0.36$

$sin^2(A)=0.64$

$sin(A)=0.8$

step 3

Find tan(A)

we know that

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

substitute the values

$tan(A)=\frac{0.8}{-0.6}$

Simplify

$tan(A)=-\frac{4}{3}$