Question

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Answer 1

QUESTION 1

The hypotenuse of the right triangle is

$24$ units

The side adjacent to the given angle is 21 units.

According to the Pythagoras' Theorem,

The length of the hypotenuse square is equal to the sum of the squares of the two shorter legs.

Let the second shorter side(opposite side) be $x$ units.

Then;

$x^2+21^2=24^2$

$x^2+441=576$

$x^2=576-441$

$x^2=135$

Take positive square root

$x=\sqrt{135}$

$x=3\sqrt{15}$

Recall the mnemonics: SOH CAH TOA

$\sin(\theta)=\frac{Opposite}{Hypotenuse}$

$\sin(\theta)=\frac{3\sqrt{15}}{24}$

Simplify:

$\sin(\theta)=\frac{\sqrt{15}}{8}$

b)  The co-secant ratio is the reciprocal sine ratio.

This implies that;

$\csc(\theta)=\frac{1}{\sin(\theta)}$

$\Rightarrow \csc(\theta)=\frac{1}{\frac{\sqrt{15}}{8}}$

$\Rightarrow \csc(\theta)=\frac{8}{\sqrt{15}}$

Rationalize the denominator;

$\Rightarrow \csc(\theta)=\frac{8\sqrt{15}}{15}$

c) The cosine ratio is

$\cos(\theta)=\frac{Adjacent}{Hypotenuse}$

$\cos(\theta)=\frac{21}{24}$

Simplify;

$\cos(\theta)=\frac{7}{8}$

d) The secant ratio is the reciprocal of the cosine ratio;

$\sec(\theta)=\frac{1}{\frac{7}{8} }$

$\sec(\theta)=\frac{8}{7}$

e) Recall and use the mnemonics TOA

$\tan(\theta)=\frac{Opposite}{Adjacent}$

$\tan(\theta)=\frac{3\sqrt{15}}{21}$

$\tan(\theta)=\frac{\sqrt{15}}{7}$

f) The co-tangent ratio is the reciprocal of the tangent

$\cot(\theta)=\frac{7}{\sqrt{15}}$

Rationalize the denominator;

$\cot(\theta)=\frac{7\sqrt{15}}{15}$

QUESTION 2

It was given that;

$\sin(\theta)=\frac{\sqrt{3}}{5}$

This means that $Opp=\sqrt{3}$ units and $Hyp=5$ units.

We apply the Pythagoras Theorem, to obtain;

$Adj^2+Opp^2=Hyp^2$

$Adj^2+(\sqrt{3})^2=5^2$

$Adj^2+3=25$

$Adj^2=25-3$

$Adj^2=22$

$Adj=\sqrt{22}$

a)  The co-secant ratio is the reciprocal sine ratio.

This implies that;

$\csc(\theta)=\frac{1}{\sin(\theta)}$

$\Rightarrow \csc(\theta)=\frac{1}{\frac{\sqrt{3}}{5}}$

$\Rightarrow \csc(\theta)=\frac{5}{\sqrt{3}}$

Rationalize the denominator;

$\Rightarrow \csc(\theta)=\frac{5\sqrt{3}}{3}$

b) The cosine ratio is

$\cos(\theta)=\frac{Adjacent}{Hypotenuse}$

$\cos(\theta)=\frac{\sqrt{22}}{5}$

c) The secant ratio is the reciprocal of the cosine ratio;

$\sec(\theta)=\frac{1}{\frac{\sqrt{22}}{5} }$

$\sec(\theta)=\frac{5}{\sqrt{22}}$

Rationalize the denominator;

$\sec(\theta)=\frac{5\sqrt{22}}{22}$

d) Recall and use the mnemonics TOA

$\tan(\theta)=\frac{Opposite}{Adjacent}$

$\tan(\theta)=\frac{\sqrt{3}}{\sqrt{22}}$

$\tan(\theta)=\frac{\sqrt{66}}{22}$

e) The co-tangent ratio is the reciprocal of the tangent

$\cot(\theta)=\frac{\sqrt{22}}{\sqrt{3}}$

Rationalize the denominator;

$\cot(\theta)=\frac{\sqrt{66}}{3}$

QUESTION 3

We  use the tangent ratio to find the opposite side.

i) $\tan(25\degree)=\frac{Opp}{34}$

This implies that;

$Opp=34\tan(25\degree)$

$Opp=15.9$  to the nearest tenth.

ii) We use the cosine ratio to find the hypotenuse;

$\cos(25\degree)=\frac{34}{Hypp}$

$Hypp=\frac{34}{\cos(25\degree)}$

$Hypp=37.5/tex] units to the nearest tenth

QUESTION 4

i) We use the tangent ratio to find the opposite side length.

[tex]\tan(48\degree)=\frac{Opp}{17}$

$Opp=17\tan(48\degree)$

$Opp=18.9$ units to the nearest tenth

ii) We use the cosine ratio to find the  hypotenuse;

$\cos(48\degree)=\frac{17}{Hypp}$

$Hypp=\frac{17}{\cos(48\degree)}$

$Hypp=25.40$ units to the nearest tenth.

QUESTION 5

We use the sine ratio to find; $\theta$

$\sin(\theta)=\frac{17}{18}$

$\theta=\sin^{-1}(\frac{17}{18})$

$\theta=70.8\degree$

QUESTION 6

We use the tangent ratio to find the value of $\theta$

$\tan(\theta)=\frac{31}{42}$

$\theta=\tan^{-1}(\frac{31}{42})$

$\theta=36.4\degree$

Answer 2

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