Question

in all problems below it’s assumed that the angle a is acute. problem 1. given sina = $\frac{4}{5}$ find cos a and tan a problem 2. given cosa = $\frac{sqrt{7}}{4}$ find sin a and tan a problem 3. given sina = 0.6 find cos a and tan a problem 4. given cosx = $\frac{12}{13}$ find the value of the expression 2cosx + 3sinx + 12tanx - 1

Explanation:

Problem 1:

Step1: Use Pythagorean identity

We know that $\sin^{2}A+\cos^{2}A = 1$. Given $\sin A=\frac{4}{5}$, then $\cos^{2}A=1 - \sin^{2}A$. So, $\cos^{2}A=1-(\frac{4}{5})^{2}=1-\frac{16}{25}=\frac{9}{25}$. Since $A$ is acute, $\cos A=\frac{3}{5}$.

Step2: Calculate tangent

$\tan A=\frac{\sin A}{\cos A}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}$

Problem 2:

Step1: Use Pythagorean identity

Since $\sin^{2}A+\cos^{2}A = 1$ and $\cos A=\frac{\sqrt{7}}{4}$, then $\sin^{2}A=1 - \cos^{2}A=1 - (\frac{\sqrt{7}}{4})^{2}=1-\frac{7}{16}=\frac{9}{16}$. Since $A$ is acute, $\sin A=\frac{3}{4}$.

Step2: Calculate tangent

$\tan A=\frac{\sin A}{\cos A}=\frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}}=\frac{3}{\sqrt{7}}=\frac{3\sqrt{7}}{7}$

Problem 3:

Step1: Use Pythagorean identity

Given $\sin A = 0.6=\frac{3}{5}$, then $\cos^{2}A=1-\sin^{2}A=1 - (\frac{3}{5})^{2}=1-\frac{9}{25}=\frac{16}{25}$. Since $A$ is acute, $\cos A=\frac{4}{5}$.

Step2: Calculate tangent

$\tan A=\frac{\sin A}{\cos A}=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$

Problem 4:

Step1: Use Pythagorean identity

Since $\sin^{2}x+\cos^{2}x = 1$ and $\cos x=\frac{12}{13}$, then $\sin^{2}x=1-\cos^{2}x=1 - (\frac{12}{13})^{2}=1-\frac{144}{169}=\frac{25}{169}$. Since $x$ is acute, $\sin x=\frac{5}{13}$.

Step2: Calculate tangent

$\tan x=\frac{\sin x}{\cos x}=\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12}$

Step3: Evaluate the expression

$2\cos x+3\sin x + 12\tan x-1=2\times\frac{12}{13}+3\times\frac{5}{13}+12\times\frac{5}{12}-1$
$=\frac{24}{13}+\frac{15}{13}+5 - 1$
$=\frac{24 + 15}{13}+4$
$=\frac{39}{13}+4$
$=3 + 4=7$

Answer:

Problem 1: $\cos A=\frac{3}{5}$, $\tan A=\frac{4}{3}$
Problem 2: $\sin A=\frac{3}{4}$, $\tan A=\frac{3\sqrt{7}}{7}$
Problem 3: $\cos A=\frac{4}{5}$, $\tan A=\frac{3}{4}$
Problem 4: $7$