Question

In triangle DOC, cos(D) = 20/29. What is the value of tan(D)?

Answer 1

Answer:

$tan(D) = \frac{21}{20}$

Step-by-step explanation:

Step(i):-

Given in triangle DOC

Cos(D) = 20/29

now

$Sec(D) = \frac{1}{Cos(D)} = \frac{1}{\frac{20}{29} } = \frac{29}{20}$

We Know that trigonometry formulas

$sec^{2}(D) -tan^{2} (D) =1$

$tan^{2} (D) = sec^{2} (D) -1$

           = $(\frac{29}{20} )^{2} -1$

          = $\frac{841 -400}{400} = \frac{441}{400}$

$tan^2(D) = \frac{441}{400}$

$tan(D) = \sqrt{\frac{441}{400} } = \frac{21}{20}$

$tan(D) = \frac{21}{20}$