Using Trigonometry and Pythagoras Together

Worked example

1. 1

   $\text{drop a height from each top corner}$ — This cuts off a right-angled triangle at each end. Its hypotenuse is the 10 cm slanted side.

2. 2

   $h=10\sin 55^{\circ }$ — The height is opposite the 55° angle and 10 is the hypotenuse, so use sin.

3. 3

   $h=8.192\dots$ — Work it out and keep the full value on your calculator for the area.

4. 4

   $e=10\cos 55^{\circ }$ — Each end piece is next to the 55° angle, with 10 as the hypotenuse, so use cos.

5. 5

   $e=5.736\dots$ — Work it out. This is one end piece only.

6. 6

   $2\times 5.736=11.472$ — There is an end piece at both ends of the trapezium, so double it.

7. 7

   $\text{top}=30-11.472=18.528$ — The top side is what is left of the 30 cm bottom after both end pieces are taken off.

8. 8

   $A=\frac{1}{2}(18.528+30)\times 8.192$ — Area of a trapezium: half the sum of the two parallel sides, times the height.

9. 9

   $A=198.7\dots =199{\text{ cm}}^2$ — Round to 3 significant figures and give the unit.

Worked example

1. 1

   $\angle ACB=90^{\circ }$ — AB is a diameter, so the angle at C on the curve is a right angle. The reason is: angle in a semicircle.

2. 2

   $\cos 40^{\circ }=\frac{14}{AB}$ — AC = 14 is next to the 40° angle and AB is the hypotenuse, so use cos.

3. 3

   $AB=14÷\cos 40^{\circ }=18.28$ — AB is the unknown on the bottom, so divide. This is the diameter.

4. 4

   $\text{arc}=\frac{1}{2}\pi \times 18.28=28.71$ — The curved edge is half a circumference, which is half of π times the diameter.

5. 5

   $28.71+18.28=47.0\text{ cm}$ — The perimeter of a semicircle is the curved arc plus the straight diameter. Round to 3 significant figures.

Worked example

1. 1

   $\angle OMR=90^{\circ }$ — A line from the centre to the midpoint of a chord meets it at a right angle. That gives a right-angled triangle $ROM$.

2. 2

   $RM=16÷2=8$ — $M$ is the midpoint, so $RM$ is half of the chord.

3. 3

   $\sin 38^{\circ }=\frac{8}{r}$ — $RM$ is opposite the 38° angle and $OR$ is a radius, the hypotenuse, so use sin.

4. 4

   $r=8÷\sin 38^{\circ }=12.99\dots$ — The radius is the unknown on the bottom, so divide. Keep the full value.

5. 5

   $A=\pi r^2=\pi \times 12.99{\dots }^2$ — Area of a circle uses the radius, not the diameter.

6. 6

   $A=530{\text{ cm}}^2$ — Round to 3 significant figures and give the unit.

Worked example

1. 1

   $\tan 58^{\circ }=\frac{9}{AD}$ — In the right-angled triangle $ABD$, 9 is opposite the 58° angle and $AD$ is next to it, so use tan.

2. 2

   $AD=9÷\tan 58^{\circ }=5.624$ — $AD$ is the unknown on the bottom, so divide.

3. 3

   $A{B}^2=9^2+{5.624}^2$ — Now two sides of the same triangle are known and no angle is needed, so use Pythagoras for the slanted side $AB$.

4. 4

   $AB=\sqrt{81+31.63}=10.61$ — Add the squares, then take the square root.

5. 5

   $AC=2\times 5.624=11.248$ — $D$ is the middle of $AC$, so the base is twice $AD$.

6. 6

   $P=2\times 10.61+11.248$ — The triangle is isosceles, so $AB$ and $BC$ are equal. Add the two slanted sides and the base.

7. 7

   $P=32.47\dots =32.5\text{ cm}$ — Round to 3 significant figures and give the unit.

Worked example

1. 1

   $QR=12\cos 35^{\circ }$ — In triangle $QRS$, $QR$ is next to the 35° angle and $QS=12$ is the hypotenuse, so use cos. $QR$ is the side both triangles share.

2. 2

   $QR=9.830\dots$ — Work it out and keep the full value.

3. 3

   $\angle PQR=35^{\circ }+20^{\circ }=55^{\circ }$ — In the big triangle $PQR$ the two angles at $Q$ sit side by side, so add them.

4. 4

   $PR=9.830\times \tan 55^{\circ }=14.04$ — $PR$ is opposite the 55° angle and $QR$ is next to it, so use tan. $PR$ is the unknown on top, so multiply.

5. 5

   $RS=12\sin 35^{\circ }=6.883$ — Back in triangle $QRS$, $RS$ is opposite the 35° angle and 12 is the hypotenuse, so use sin.

6. 6

   $PS=14.04-6.883=7.16\text{ cm}$ — $S$ lies between $P$ and $R$, so $PS$ is the whole of $PR$ take away the part $RS$.