2. Trigonometry (SOHCAHTOA) - Cardiff Tutor Company

Question 1

LEVEL 6

$ABC$ is a right angled triangle.

$AC = 12$ cm, $\angle CAB = 58\degree$

Calculate the length of $BC$ , labelled $x$ .

Select the correct answer from the list below:

A: $8.6$

B: $10.2$

C: $9.4$

D: $11.3$

CORRECT ANSWER: B: $10.2$

WORKED SOLUTION:

To find $x$ we can use

$\text{sin}\theta=$ $\dfrac{\text{opposite}}{\text{hypotenuse}}$

In this instance we are given the angle and the hypotenuse, so

$\text{sin}(58\degree)= \dfrac{x}{12}$

So we have to rearrange the formula to calculate the value of the opposite length by multiplying both sides of the equation by $12$

$x=12 \, \text{sin} (58\degree)=10.2$ cm

Question 2

LEVEL 6

$ABC$ is a right angled triangle.

$AC = 11$ cm, $CB = 8$ cm

Calculate the angle $CAB$ , labelled $x$ .

Select the correct answer from the list below:

A: $31.6\degree$

B: $52.4\degree$

C: $46.7\degree$

D: $38.5\degree$

CORRECT ANSWER: C: $46.7\degree$

WORKED SOLUTION:

To find $\angle x$ we can use

$\text{sin} \theta=\dfrac{\text{opposite}}{\text{hypotenuse}}$

In this instance we are given the hypotenuse and the opposite lengths, so

$\text{sin} (x\degree)=\dfrac{8}{11}$

In order to find $\angle x$ we have to calculate the inverse sine of this fraction,

$x=\text{sin}^{-1}(\dfrac{8}{11})=46.7\degree$

Question 3

LEVEL 6

$ABC$ is a right angled triangle.

$AB = 9$ cm, $\angle CAB = 36\degree$

Calculate the length of $AC$ , labelled $x$ .

Select the correct answer from the list below:

A: $11.1$ cm

B: $14.2$ cm

C: $12.8$ cm

D: $10.3$ cm

CORRECT ANSWER: A: $11.1$ cm

WORKED SOLUTION:

To find $x$ we can use

$\text{cos} \theta= \dfrac{\text{adjacent}}{\text{hypotenuse}}$

In this case we are given the adjacent length and an angle.

$\text{cos} (36\degree)=$ $\dfrac{9}{x}$

To find $x$ we have to rearrange the formula so that $x$ is the subject

$x=\dfrac{9}{\text{cos} (36\degree)} =11.1$ cm

Question 4

LEVEL 6

$ABC$ is a right angled triangle.

$AB = 10$ cm, $\angle CAB = 50\degree$

Calculate the length of $BC$ to $1$ , labelled $x$ .

Select the correct answer from the list below:

A: $11.9$ cm

B: $11.5$ cm

C: $12.9$ cm

D: $12.5$ cm

CORRECT ANSWER: A: $11.9$ cm

WORKED SOLUTION:

To find $x$ we can use

$\text{tan} \theta=\dfrac{\text{opposite}}{\text{adjacent}}$

In this case we are given an angle and the adjacent length and asked to find the opposite length.

$\text{tan} (50\degree)= \dfrac{x}{10}$

We can calculate the value of $x$ by multiplying both sides of the equation by $10$

$x= 10 \, \text{tan} (50\degree) =11.9$ cm

Question 5

LEVEL 6

A flag pole, $P$ , is $4.3$ m West of the scout hut, $H$ .

A camp fire, $F$ , is $5.5$ m North of the flag pole.

Calculate the size of the angle $y$ .

Select the correct answer from the list below:

A: $47\degree$

B: $52\degree$

C: $56\degree$

D: $49\degree$

CORRECT ANSWER: B: $52\degree$

WORKED SOLUTION:

To find $\angle y\degree$ we can use

$\text{tan} \theta = \dfrac{\text{opposite}}{\text{adjacent}}$

In this instance we are given the opposite and the adjacent side lengths so,

$\text{tan} (y\degree)=\dfrac{5.5}{4.3}$

Taking the inverse tangent of both sides of the equation gives us the value of $y$

$y = \text{tan}^{-1}(\dfrac{5.5}{4.3})=52\degree$

Question 6

LEVEL 6

Find the length of the missing side, labelled with a $?$ .

Select the correct answer from the list below:

A: $5.76$ m

B: $8.31$ m

C: $7.37$ m

D: $9.02$ m

CORRECT ANSWER: D: $9.02$ m

WORKED SOLUTION:

First we have to find the height by considering the green triangle,

$h=4.04\times \text{cos} (9.9\degree)=3.98$ m

Then we can work out the missing length, labelled $?$ ,

$\text{Length} =\dfrac{3.98}{\text{sin} (26.18\degree)}=9.02$ m

Question 7

LEVEL 6

Find the missing angle, labelled with a $?$ .

Select the correct answer from the list below:

A: $83.95\degree$

B: $86.90\degree$

C: $88.00\degree$

D: $87.99\degree$

CORRECT ANSWER: D: $87.99\degree$

WORKED SOLUTION:

First we have to find the height by considering the orange triangle,

$h=51\times \text{sin}(6.72\degree)=5.97$ m

Then we can work out the missing angle, labelled $?$

$\text{Angle} = \text{tan}^{-1}(\dfrac{5.97}{0.21})=87.99\degree$