Sine, Cosine Rule & Area of Triangles Flashcards

All 26 cards in this deck

State the sine rule in the form used to find a missing side.

$\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

Each side is paired with the angle opposite it.

State the sine rule in the form used to find a missing angle.

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$

What are the steps to find a missing side using the sine rule?

e.g. find $x$, the side opposite $40^\circ$, when the side opposite $80^\circ$ is $10$ cm

1. Pair each side with its opposite angle: $\dfrac{x}{\sin 40^\circ}=\dfrac{10}{\sin 80^\circ}$
2. Multiply by $\sin 40^\circ$: $x=\dfrac{10\sin 40^\circ}{\sin 80^\circ}$
3. Evaluate: $x=6.53$ cm (3 s.f.)

What are the steps to find a missing angle using the sine rule?

e.g. find angle $B$, opposite $7$ cm, when the side opposite $30^\circ$ is $10$ cm

1. Use $\dfrac{\sin B}{b}=\dfrac{\sin A}{a}$: $\dfrac{\sin B}{7}=\dfrac{\sin 30^\circ}{10}$
2. Rearrange: $\sin B=\dfrac{7\sin 30^\circ}{10}=0.35$
3. Inverse sine: $B=\sin^{-1}(0.35)=20.5^\circ$ (1 d.p.)

True or false? In the sine rule, the side $a$ is the side next to angle $A$.

False.

$a$ is the side opposite angle $A$.

The sine rule gives $\sin x = 0.6$, and the diagram shows $x$ is obtuse. How do you obtain the obtuse angle?

Work out $180^\circ-\sin^{-1}(0.6)$.

The calculator only gives the acute solution.

State the cosine rule for finding a missing side.

$a^2=b^2+c^2-2bc\cos A$

$A$ is the angle between sides $b$ and $c$, opposite side $a$.

State the cosine rule rearranged to find a missing angle.

$\cos A=\dfrac{b^2+c^2-a^2}{2bc}$

What are the steps to find a side using the cosine rule?

e.g. sides $8$ cm and $11$ cm with an included angle of $72^\circ$

1. Substitute: $a^2=8^2+11^2-2\times 8\times 11\times\cos 72^\circ$
2. Correct order of operations: $64+121-54.4=130.6$
3. Square root: $a=11.4$ cm (3 s.f.)

What are the steps to find an angle using the cosine rule?

e.g. find the angle opposite the $7$ cm side in a triangle with sides $5$ cm, $8$ cm, $7$ cm

1. Substitute into $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$: $\cos A=\dfrac{5^2+8^2-7^2}{2\times5\times8}$
2. Simplify: $\cos A=\dfrac{40}{80}=0.5$
3. Inverse cosine: $A=60^\circ$

True or false? In $a^2=b^2+c^2-2bc\cos A$ you may work out $b^2+c^2-2bc$ first and then multiply by $\cos A$.

False.

You must multiply $2bc$ by $\cos A$ first, then subtract that from $b^2+c^2$.

What are the steps to use the cosine rule in a bearings problem?

e.g. a ship sails $8$ km on a bearing of $060^\circ$, then $5$ km on a bearing of $150^\circ$; find its distance from the start

1. Sketch the journey with a north line at each turning point.
2. Use angle facts to find the interior angle of the triangle: $90^\circ$
3. Apply the cosine rule: $d^2=8^2+5^2-2\times8\times5\cos 90^\circ$, so $d=9.43$ km (3 s.f.)

What is the formula for the area of any triangle using two sides and an angle?

Area $=\tfrac12 ab\sin C$

$C$ is the angle included between sides $a$ and $b$.

True or false? $\tfrac12 ab\sin C$ works with any two sides and any angle of the triangle.

False.

The angle must be the one included between the two sides used.

What are the steps to find the area of a triangle from two sides and the included angle?

e.g. sides $8$ cm and $11$ cm with an included angle of $72^\circ$

1. Write the formula: Area $=\tfrac12 ab\sin C$
2. Substitute: $0.5\times 8\times 11\times\sin 72^\circ$
3. Evaluate: $41.8$ cm$^2$ (3 s.f.)

What are the steps to find an included angle when the area is given?

e.g. a triangle with sides $8$ cm and $10$ cm has area $20$ cm$^2$

1. Substitute into $\tfrac12 ab\sin C=$ Area: $0.5\times 8\times 10\times\sin C=20$
2. Rearrange: $\sin C=\dfrac{20}{40}=0.5$
3. Inverse sine: $C=30^\circ$ (or $150^\circ$ if obtuse)

What are the steps to find $x$ when the area formula gives an equation in $x$?

e.g. sides $(x+3)$ m and $(2x-1)$ m, included angle $45^\circ$, area $6\sqrt2$ m$^2$

1. Set up the equation: $\tfrac12(x+3)(2x-1)\sin 45^\circ=6\sqrt2$
2. Expand and rearrange: $2x^2+5x-27=0$
3. Solve with the quadratic formula: $x=2.63$ (3 s.f.)

True or false? $\tfrac12 ab\sin C$ cannot be used when the included angle is obtuse.

False.

The sine of an obtuse angle is positive, so the formula still gives a positive area.

Which rule do you use to find the third side when you know two sides and the angle between them?

The cosine rule, $a^2=b^2+c^2-2bc\cos A$.

Which rule do you use to find an angle when you know all three sides?

The cosine rule, $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$.

Which rule do you use when you know a side and the angle opposite it, plus one more side or angle?

The sine rule.

Which methods should you use in a triangle that contains a right angle?

Pythagoras' theorem and SOHCAHTOA.

They are quicker, though the sine and cosine rules would still work.

True or false? The cosine rule can find a missing side when you know two sides and a non-included angle.

False.

The angle must be the included angle; otherwise start with the sine rule.

What are the steps to solve a problem across two adjoining triangles?

e.g. triangles $ABC$ and $ADC$ share side $AC$; area of $ADC$ is known, $AB$ is required

1. Use the given information in the first triangle (here $\tfrac12 ab\sin C=$ area) to find its missing side.
2. Use the cosine (or sine) rule in that triangle to find the shared side $AC$.
3. Use the sine (or cosine) rule in the second triangle with $AC$ to find $AB$.

True or false? If the cosine rule gives a negative value for $\cos A$, then angle $A$ is obtuse.

True. $\cos$ is negative for angles between $90^\circ$ and $180^\circ$, so $A$ is obtuse.

What are the steps to find the distance from the start of a two-leg journey given as bearings?

e.g. a boat sails $5$ km on a bearing of $050^\circ$, then $7$ km on a bearing of $110^\circ$

1. Sketch the journey and find the angle inside the triangle at the turning point: $180^\circ-(110^\circ-50^\circ)=120^\circ$
2. Two sides and the included angle, so use the cosine rule: $d^2=5^2+7^2-2\times5\times7\times\cos120^\circ=109$
3. Square root: $d\approx10.4$ km