A right-angled triangle has a 40° angle at the bottom right. The side opposite this angle is 12 cm, the bottom side is 18 cm, and the sloping side is the longest side. Name the opposite, adjacent and hypotenuse.

1. The hypotenuse is the longest side, so the sloping side is the hypotenuse.

2. The side directly across from the 40° angle is the opposite side, so 12 cm is the opposite.

3. The side next to the 40° angle that is not the hypotenuse is the adjacent side, so 18 cm is the adjacent.

In a right-angled triangle, the angle is 35°. The side opposite the angle is 20 cm and the hypotenuse is $x$ cm. Work out which trig ratio to use.

1. Label the sides relative to the 35° angle: 20 cm is opposite and $x$ is the hypotenuse.

2. The sides involved are opposite and hypotenuse.

3. Opposite and hypotenuse go with SOH, so use sine:

   $$\sin \theta =\frac{opposite}{hypotenuse}$$

A right-angled triangle has an angle of 35°. The side opposite this angle is 20 cm. The hypotenuse is $x$ cm. Find $x$.

1. Label the sides: 20 cm is opposite and $x$ is the hypotenuse.

2. Choose the ratio. Opposite and hypotenuse means sine:

   $$\sin 35^{\circ }=\frac{20}{x}$$

3. Rearrange by multiplying both sides by $x$:

   $$x\sin 35^{\circ }=20$$

4. Divide by $\sin 35^{\circ }$:

   $$x=\frac{20}{\sin 35^{\circ }}$$

5. Use your calculator:

   $$x=34.9cm$$

A right-angled triangle has an angle of 42°. The side opposite this angle is 11 cm. The adjacent side is $x$ cm. Find $x$.

1. Label the sides: 11 cm is opposite and $x$ is adjacent.

2. Choose the ratio. Opposite and adjacent means tangent:

   $$\tan 42^{\circ }=\frac{11}{x}$$

3. Multiply both sides by $x$:

   $$x\tan 42^{\circ }=11$$

4. Divide by $\tan 42^{\circ }$:

   $$x=\frac{11}{\tan 42^{\circ }}$$

5. Calculate:

   $$x=12.2cm$$

A right-angled triangle has a side next to angle $x$ of 16 cm and a hypotenuse of 24 cm. Find angle $x$.

1. Label the sides relative to angle $x$: 16 cm is adjacent and 24 cm is the hypotenuse.

2. Choose the ratio. Adjacent and hypotenuse means cosine:

   $$\cos x=\frac{16}{24}$$

3. Use inverse cosine:

   $$x={\cos }^{-1}\left(\frac{16}{24}\right)$$

4. Calculate:

   $$x=48.2^{\circ }$$

A right-angled triangle has opposite side 11 cm and adjacent side 15 cm relative to angle $x$. Find $x$.

1. The sides involved are opposite and adjacent.

2. Opposite and adjacent means tangent:

   $$\tan x=\frac{11}{15}$$

3. Use inverse tangent:

   $$x={\tan }^{-1}\left(\frac{11}{15}\right)$$

4. Calculate:

   $$x=36.3^{\circ }$$

Triangle $ABC$ is right-angled at $B$. The side $BC$ is 20 cm and angle $C$ is 35°. Calculate $AB$.

1. Focus on angle $C$.

2. Side $AB$ is opposite angle $C$, and side $BC$ is adjacent to angle $C$.

3. Opposite and adjacent means tangent:

   $$\tan 35^{\circ }=\frac{AB}{20}$$

4. Multiply by 20:

   $$AB=20\tan 35^{\circ }$$

5. Calculate:

   $$AB=14.0cm$$

A right-angled trapezium has vertical sides 4 m and 6 m, and the distance between them is 15 m. Find the angle between the shorter vertical side and the sloping top. Give your answer to 1 decimal place.

1. The height difference between the two vertical sides is:

   $$6-4=2$$

2. The horizontal distance across the shape is 15 m, so the small right triangle has opposite side 2 m and adjacent side 15 m.

3. Use tangent:

   $$\tan \theta =\frac{2}{15}$$

4. Find the small angle above the horizontal:

   $$\theta ={\tan }^{-1}\left(\frac{2}{15}\right)$$

5. Calculate:

   $$\theta =7.6^{\circ }$$

6. The required angle with the vertical side is the complement of this angle:

   $$90^{\circ }-7.6^{\circ }=82.4^{\circ }$$

A right-angled triangle has shorter sides 20 cm and 16 cm. The hypotenuse is $x$ cm. Find $x$ to 1 decimal place.

1. This question is asking for the hypotenuse and gives the two shorter sides, so Pythagoras is the quickest method.

2. Use $a^2+b^2=c^2$:

   $$x^2=20^2+16^2$$

3. Work out the squares:

   $$x^2=400+256=656$$

4. Square root:

   $$x=\sqrt{656}$$

5. Round to 1 decimal place:

   $$x=25.6cm$$