Trigonometry Questions, Revision and Worksheets

In this case, the two sides we’re concerned with are the hypotenuse and the side adjacent to the angle given.   Therefore, we want the ‘CAH’ part of ‘SOH CAH TOA’ where A=35,H=p, and the angle is 43\degree:

\cos(43°)=\dfrac{35}{p}

Next we need to solve for p. Multiplying both side by p to get:

p\times\cos(43°)=35

Then, as \cos(43°) is just a number, we can we can divide both sides by \cos(43°):

p=\dfrac{35}{\cos(43°)}

Finally, putting this into the calculator we get:

p=47.85646...

p=47.9 m (3 sf)

The two sides we’re working with are the hypotenuse and the opposite side, therefore, we want the ‘SOH’ part of ‘SOH CAH TOA’, where  O=13,H=15, and the angle is q°:

\sin(q)=\dfrac{13}{15}

To find q, we have to apply the inverse sine function to both sides. It cancels out the sin on the left-hand side and we get:

q=\sin^{-1}\left(\dfrac{13}{15}\right)

q=60.073565...

q=60.1\degree (1 dp).

According to SOH CAH TOA, the sin of w must be equal to the opposite side divided by the hypotenuse. The opposite side is given to us: 2, but the hypotenuse is not. Since we have a right-angled triangle, we can use Pythagoras to find the hypotenuse. If the hypotenuse is c,  then a and b are both 2, so the equation a^2+b^2=c^2 becomes:

c^2=2^2+2^2=4+4=8

Square rooting both sides, we get:

c=\sqrt{8}=2\sqrt{2}

Now we have the hypotenuse, we can use ‘SOH’:

\sin(w)=\dfrac{2}{2\sqrt{2}}

Notice that there is a 2 on the top and bottom that can cancel:

\sin(w)=\dfrac{2}{2\sqrt{2}}=\dfrac{1}{\sqrt{2}}

In this case, the two sides we’re working with are the hypotenuse, and the side opposite to the angle given.   Therefore, we want the ‘SOH’ part of ‘SOH CAH TOA’ where O=CB,H=12, and the angle is 30\degree:

\sin(30°)=\dfrac{CB}{12}

Next we can solve this by multiplying both side by 12 to get:

12\sin(30°)=CB

Finally, putting this into the calculator we get:

CB=6.00 cm (3 sf)

The two sides we’re working with are the adjacent and the opposite. Therefore, we want the ‘TOA’ part of ‘SOH CAH TOA’, where  O=4,A=7, and the angle is x:

\tan(x)=\dfrac{4}{7}

To find x, we have to apply the inverse tangent function to both sides. It cancels out the tan on the left-hand side and we get:

x=\tan^{-1}\left(\dfrac{4}{7}\right)=29.7448813...

x=29.7° (1 dp).