Question

problem 1. given the following right - angled triangle, find:
a) sina, cosa, tana, cota
b) sin b, cosb, tanb, cotb
problem 2. simplify
5sin²x + 6x + 5cos²x - 5
problem 3. simplify
7 - cosx·tanx - 2sinx + tanx·cotx
problem 4. given cosa = 2√6/7 find sin a, tan a and cota
problem 5. given the triangle below find the value of the expression
10sina - 5 cosa + 3 tana + 6

Explanation:

Step1: Recall trigonometric ratios

For a right - angled triangle, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$, $\cot\theta=\frac{\text{adjacent}}{\text{opposite}}$

Part a)

• For $\sin A$:

Opposite side to angle $A$ is $BC = 2$, hypotenuse $AB=\sqrt{5}$
$\sin A=\frac{BC}{AB}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$

• For $\cos A$:

Adjacent side to angle $A$ is $AC = 1$, hypotenuse $AB=\sqrt{5}$
$\cos A=\frac{AC}{AB}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$

• For $\tan A$:

Opposite side to angle $A$ is $BC = 2$, adjacent side to angle $A$ is $AC = 1$
$\tan A=\frac{BC}{AC}=\frac{2}{1} = 2$

• For $\cot A$:

Adjacent side to angle $A$ is $AC = 1$, opposite side to angle $A$ is $BC = 2$
$\cot A=\frac{AC}{BC}=\frac{1}{2}$

Part b)

• For $\sin B$:

Opposite side to angle $B$ is $AC = 1$, hypotenuse $AB=\sqrt{5}$
$\sin B=\frac{AC}{AB}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$

• For $\cos B$:

Adjacent side to angle $B$ is $BC = 2$, hypotenuse $AB=\sqrt{5}$
$\cos B=\frac{BC}{AB}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$

• For $\tan B$:

Opposite side to angle $B$ is $AC = 1$, adjacent side to angle $B$ is $BC = 2$
$\tan B=\frac{AC}{BC}=\frac{1}{2}$

• For $\cot B$:

Adjacent side to angle $B$ is $BC = 2$, opposite side to angle $B$ is $AC = 1$
$\cot B=\frac{BC}{AC}=2$

Answer:

a) $\sin A=\frac{2\sqrt{5}}{5}$, $\cos A=\frac{\sqrt{5}}{5}$, $\tan A = 2$, $\cot A=\frac{1}{2}$
b) $\sin B=\frac{\sqrt{5}}{5}$, $\cos B=\frac{2\sqrt{5}}{5}$, $\tan B=\frac{1}{2}$, $\cot B = 2$