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{A}{H}=\dfrac{35}{p}

Now, we must rearrange this by multiplying both side by p to get,

p\times\cos(43)=35

Then, remember that \cos(43) is just a number, so we can divide both sides by “\cos(43)”, 

p=\dfrac{35}{\cos(43)}

Finally, putting this into the calculator we get:

p=47.85646...=47.9\text{ m (3 s.f.)}

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

\sin(q)=\dfrac{O}{H}=\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)=60.1\degree\text{ (1 d.p.)}

Since two sides of this triangle are the same length, it must be an isosceles triangle. In an isosceles triangle, we must have two angles the same and as the third angle in this instance is 90 degrees then, 

w\degree+w\degree+90\degree=180\degree

Subtracting 90 degrees from both sides we find,

2w=90

Then, dividing both sides by 2 gives us, w=45, thus: 

\sin(w)=\sin(45)=\dfrac{1}{\sqrt{2}}

ALTERNATIVE METHOD:

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. However, we can find it using Pythagoras, since this is a right-angled triangle. 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 know the hypotenuse, we get,

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

Notice that there is a 2 on top and bottom that can cancel, and so we find,

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

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

\sin(30)=\dfrac{O}{H}=\dfrac{x}{12}

Now, we must rearrange this by multiplying both side by 12 to get,

12\sin(30)=x

Finally, putting this into the calculator we get:

x=6 \text{ cm (3 s.f.)}

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

\tan(x)=\dfrac{O}{A}=\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\degree=\tan^{-1}\left(\dfrac{4}{7}\right)=29.7\degree\text{ (1 d.p.)}