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Here we will learn how to use SOHCAHTOA and trigonometry to find unknown sides and angles in right angled triangles. You’ll learn how to label the sides of right-angled triangles, what sin, cos and tan are, what their inverses are (
Look out for the trigonometry practice problems, worksheets and exam questions at the end.
SOHCAHTOA is a mnemonic that gives us an easy way to remember the three main trigonometric ratios. They are sine (
We can use these to work out missing sides and angles in right-angled triangles.
SOH stands for:
CAH stands for:
TOA stands for:
We can abbreviate these to the SOHCAHTOA triangles:
The hypotenuse is the longest side of the triangle. It is opposite the right angle.
The opposite side is the side that is opposite the angle.
The adjacent side is the side that is adjacent (next to) the angle.
E.g.
We have a right-angled triangle with a side of length
Label the sides of the right-angled triangle.
2Label the side opposite the angle O (opposite)
3Label the side next to the angle A (adjacent)
Top tip: once you have labeled the hypotenuse (H) and the opposite (O), the adjacent (A) must be the only side left!
We can use trigonometry to work out the unknown sides of a right-angled triangle by using SOHCAHTOA.
Get your free SOHCAHTOA worksheet of 20+ questions and answers. Includes reasoning and applied questions.
DOWNLOAD FREEGet your free SOHCAHTOA worksheet of 20+ questions and answers. Includes reasoning and applied questions.
DOWNLOAD FREECalculate the side labeled
2Choose the trig ratio we need.
The trigonometric function that involves the opposite (O) and the hypotenuse (H) is
3Substitute the values from the triangle into the function.
4Calculate the unknown side, rearranging if necessary. We can either solve the equation or use the SOHCAHTOA triangles.
Always check that the answer is sensible:
As
Calculate the side labeled
Label the sides of the right-angled triangle that we have information about.
Circle the labels so not to confuse them with the side lengths.
Choose the trig ratio we need.
The trigonometric function that involves the adjacent (A) and the hypotenuse (H) is
Substitute the values from the triangle into the function.
Calculate the unknown side, rearranging if necessary.
We can either solve the equation or use the SOHCAHTOA triangles.
Always check that the answer is sensible:
As
Calculate the side labeled
Label the sides of the right-angled triangle that we have information about.
Circle the labels so not to confuse them with the side lengths.
Choose the trig ratio we need.
The trigonometric function that involves the opposite (O) and the adjacent (A) is
Substitute the values from the triangle into the function.
Calculate the unknown side, rearranging if necessary.
We can either solve the equation or use the SOHCAHTOA triangles.
Always check that the answer is sensible:
A good sketch tells us this answer looks reasonable.
1. Calculate the length of the side labeled x . Give your answer to 1 d.p.
7.8 cm
18.4 cm
60.4 cm
16.9 cm
Label the triangle:
We know H and we want to work out O so use sin.
\begin{aligned} \sin(\theta)&=\frac{O}{H}\\ \sin(23)&=\frac{x}{20}\\ 20 \times \sin(23) &= x\\ 7.81462257 &= x\\ x&=7.8\mathrm{cm} \end{aligned}
2. Calculate the length of the side labeled x . Give your answer to 2 d.p.
22.42 cm
37.50 cm
20.18 cm
11.15 cm
Label the triangle:
We know A and we want to find H so use cos.
\begin{aligned} \cos(\theta)&=\frac{A}{H}\\ \cos(42)&=\frac{15}{x}\\ x \times \cos(42) &=15\\ x &= \frac{15}{\cos(42)}\\ x &=20.18449094\\ x &=20.18 \mathrm{cm} \end{aligned}
3. Calculate the length of the side labeled x . Give your answer to 2 s.f.
6.0 cm
4.8 cm
11 cm
2.8 cm
Label the triangle:
We know O and we want to work out A so use tan.
\begin{aligned} \tan(\theta)&=\frac{O}{A}\\ \tan(62)&=\frac{5.3}{x}\\ x \times \tan(62) &= 5.3\\ x &=\frac{5.3}{\tan(62)}\\ x &= 2.818059988\\ x &= 2.8 \mathrm{cm} \end{aligned}
In order to work out missing angles in right-angled triangles we need use the inverse trigonometric functions:
We can find these on the calculator by pressing SHIFT and then
Step by step guide: Inverse trigonometric functions (coming soon)
We can use trigonometry to work out the unknown angles of a right-angled triangle by using SOHCAHTOA.
Calculate the angle labeled θ.
Label the sides of the right-angled triangle that we have information about.
Choose the trig ratio we need.
The trigonometric function that involves the opposite (O) and the hypotenuse (H) is
Substitute the values from the triangle into the function.
Using inverse trig functions, work out the missing angle θ.
Remember we can get
Always check that the answer is sensible: We can estimate from the diagram that θ is an acute angle.
As
Calculate the angle labeled θ.
Label the sides of the right-angled triangle that we have information about.
Choose the trig ratio we need.
The trigonometric function that involves the adjacent (A) and the hypotenuse (H) is
Substitute the values from the triangle into the function.
Using inverse trig functions, work out the missing angle θ.
Remember we can get
Always check that the answer is sensible: We can estimate from the diagram that θ is an acute angle.
As
Calculate the angle labeled θ.
Label the sides of the right-angled triangle that we have information about.
Choose the trig ratio we need.
The trigonometric function that involves the opposite (O) and the adjacent (A) is
Substitute the values from the triangle into the function.
Using inverse trig functions, work out the missing angle θ.
Remember we can get
Always check that the answer is sensible: We can estimate from the diagram that θ is an acute angle.
As
1. Calculate the size of the angle labeled \theta . Give your answer to 3 s.f.
Label the triangle:
We know O and H so use sin.
\begin{aligned} \sin(\theta)&=\frac{O}{H}\\ \sin(\theta)&=\frac{8}{11}\\ \theta&=\sin^{-1}(\frac{8}{11})\\ \theta &=46.65824177\\ \theta&=47.7^{\circ} \end{aligned}
2. Calculate the size of the angle labeled \theta . Give your answer to 2 s.f.
Label the triangle:
We know A and H so use cos.
\begin{aligned} \cos(\theta)&=\frac{A}{H}\\ \cos(\theta)&=\frac{5.4}{12.3}\\ \theta&=\cos^{-1}(\frac{5.4}{12.3})\\ \theta&=63.95835004\\ \theta&=64^{\circ} \end{aligned}
3. Calculate the size of the angle labeled \theta . Give your answer to 1 d.p.
Label the triangle:
We know O and A so use tan.
\begin{aligned} \tan(\theta)&=\frac{O}{A}\\ \tan(\theta)&=\frac{12.6}{25.2}\\ \theta &= \tan^{-1}(\frac{12.6}{25.2})\\ \theta&=26.56505118\\ \theta&= 26.6^{\circ} \end{aligned}
1. Find the size of angle a to 2.d.p.
(3 marks)
(1)
\cos(a) = \frac{6}{13}
(1)
\begin{aligned} a&=\cos^{-1}(\frac{6}{13})\\ a&=62.51^{\circ} \end{aligned}
(1)
2. Rosie is flying a kite. She holds the kite string 1m from the ground. The kite string is 10m long and the kite flies at an angle of 40^{\circ} . Calculate the height of the kite above the ground.
(3 marks)
(1)
opp=6.43m
(1)
total height:
6.43+1=7.43m(1)
3. Length AB = 15cm
Angle ABE = 51^{\circ}
Angle ADE = 32^{\circ}
Work out the length of DE.
(4 marks)
Length AE:
\begin{aligned} \tan(51)&=\frac{AE}{15}\\ 15\tan(51)&=AE \end{aligned}
(1)
AE = 18.52
(1)
Length DE:
\begin{aligned} \tan(32)&=\frac{18.52}{x}\\ x&=\frac{18.52}{tan(32)} \end{aligned}
(1)
x=29.64cm
(1)
You have now learned how to:
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