Question

Trigonometric Ratios Find

(a).$\begin{array}{l}\sin \alpha and\cos \beta ,\end{array}$

(b). $\tan \alpha and\cot \beta$, and

(c). $\tan \alpha and\cot \beta$.

Short answer

(a).

$\begin{array}{l}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\alpha }\mathbf{=}\frac{\mathbf{3}}{\sqrt{\mathbf{34}}}\mathbf{\text{and}}\\ \mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\beta }\mathbf{=}\frac{\mathbf{3}}{\sqrt{\mathbf{34}}}\mathbf{,}\mathbf{\text{forthetriangle}}\end{array}$

(b).

$\mathit{t}\mathit{a}\mathit{n}\mathit{\alpha }\mathbf{=}\frac{\mathbf{3}}{\mathbf{5}}\mathit{a}\mathit{n}\mathit{d}\mathit{c}\mathit{o}\mathit{t}\mathit{\beta }\mathbf{=}\frac{\mathbf{3}}{\mathbf{5}}\mathbf{,}\mathit{f}\mathit{o}\mathit{r}\mathit{t}\mathit{h}\mathit{e}\mathit{t}\mathit{r}\mathit{i}\mathit{a}\mathit{n}\mathit{g}\mathit{l}\mathit{e}$

(c).

$\begin{array}{l}\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{\alpha }\mathbf{=}\frac{\sqrt{\mathbf{34}}}{\mathbf{5}}\mathbf{\text{and}}\\ \mathbf{cosec}\mathbf{\beta }\mathbf{=}\frac{\sqrt{\mathbf{34}}}{\mathbf{5}}\mathbf{,}\mathbf{\text{forthetriangle}}\end{array}$

Step-by-step solution

a. Step 1. Given information.

We have$P$is perpendicular,$B$is base,$H$is hypotenuse.

Here for $\angle \alpha ,P=3B=5$

By applying Pythagoras Theorem, we get

a. Step 2. Finding the value of trigonometric values.

$$\begin{gathered} H=\sqrt{P^2+B^2} \\ \Rightarrow H=\sqrt{3^2+5^2} \\ \Rightarrow H=\sqrt{9+25} \\ \Rightarrow H=\sqrt{34} \\ \sin \alpha =\frac{P}{H} \\ =\frac{3}{\sqrt{34}} \end{gathered}$$

$$\begin{gathered} \angle \beta ,P=5 \\ B=3 \\ \cos \beta =\frac{H}{H} \\ =\frac{3}{\sqrt{34}} \\ \begin{array}{c}H=\sqrt{P^2+B^2}\\ \Rightarrow H=\sqrt{3^2+5^2}\\ \Rightarrow H=\sqrt{9+25}\\ \Rightarrow H=\sqrt{34}\\ \sin \alpha =\frac{P}{H}\\ =\frac{3}{\sqrt{34}}\\ \text{for,}\angle \beta ,P=5\\ B=3\\ \cos \beta =\frac{H}{H}\\ =\frac{3}{\sqrt{34}}\end{array} \end{gathered}$$

b. Step 1. Given information.

We have$P$ is perpendicular, $B$is base, $H$is hypotenuse.

Here for $\angle \alpha ,P=3B=5$

By applying Pythagoras Theorem, we get

b. Step 2.Finding trigonometric values.

$H=\sqrt{P^2+B^2}$

$\Rightarrow H=\sqrt{3^2+5^2}$

$\Rightarrow H=\sqrt{9+25}$

$\Rightarrow H=\sqrt{34}$

$\tan \alpha =\frac{P}{B}$

$=\frac{3}{5}$

for, $\angle \beta ,P=5$

$B=3$

$\cot \beta =\frac{B}{P}$

$=\frac{3}{5}$

c. Step 1. Given information.

We have $P$is perpendicular, $B$is base,$H$is hypotenuse.

Here for$\angle \alpha ,P=3B=5$

By applying Pythagoras Theorem, we get

c. Step 2. Finding trigonometric values.

$H=\sqrt{P^2+B^2}$

$\Rightarrow H=\sqrt{3^2+5^2}$

$\Rightarrow H=\sqrt{9+25}$

$\Rightarrow H=\sqrt{34}$

$\text{sec}\alpha =\frac{H}{B}$

for, $\angle \beta ,P=5$

$=\frac{\sqrt{34}}{5}$

$\mathrm{cosec}\beta =\frac{H}{P}$

$\Rightarrow \frac{\sqrt{34}}{5}$