TRIGONOMETRY · HL
Non-Right-Angled Triangles
Sin Rule, Cosine Rule and Area = ½ ab Sin C.
Section 1 of 9
The triangle ABC
Triangle ABC.
$\angle A = \angle BAC$
$|a| = |BC|$
Sin Rule
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$ or $\dfrac{\sin A}{a} = \dfrac{\sin B}{b}$
Cosine Rule
$a^2 = b^2 + c^2 - 2bc\cos A$
or $b^2 = a^2 + c^2 - 2ac\cos B$
or $c^2 = a^2 + b^2 - 2ab\cos C$
Area
Area $= \tfrac{1}{2}\,ab\sin C$
Section 2 of 9
Sin Rule
$a$ to 1 decimal place
Opposites full? Yes $\Rightarrow$ Sin Rule
Each known angle is joined to the side opposite it.
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$
$\dfrac{a}{\sin 50} = \dfrac{5}{\sin 70}$
$a = \dfrac{5\sin 50}{\sin 70}$
$= 4.076$
$= 4.1$
Section 3 of 9
Cosine Rule
Try to find opposite? = None
Cannot use Sin Rule.
Which is $b$ and which is $c$? Does not matter
$a^2 = b^2 + c^2 - 2bc\cos A$
$a^2 = 5^2 + 6^2 - 2(5)(6)\cos 30$
$a^2 = 25 + 36 - 60\cos 30$
$a^2 = 61 - 60\cos 30$
$a^2 = (61-60)\cos 30 = \cos 30$ (wrong — you cannot subtract before multiplying)
$a^2 = 9.03$
$a = 3.006$
$a = 3$
Section 4 of 9
Cosine Rule — find the largest angle
Find Largest angle.
Largest angle is opposite longest side.
No opposites – no angles.
$a^2 = b^2 + c^2 - 2bc\cos A$
$7^2 = 6^2 + 5^2 - 2(6)(5)\cos A$
$49 = 36 + 25 - 60\cos A$
$49 = 61 - 60\cos A$
$60\cos A = 12$
$\cos A = \dfrac{12}{60} = \dfrac{1}{5}$
$A = 78.46^\circ$
Section 5 of 9
Area
Find area.
Area $= \tfrac{1}{2}\,ab\sin C$
$\Rightarrow$ 2 lengths and angle inbetween.
Area $= \tfrac{1}{2}\,(5)(7)\sin 45$
$= 12.4$ units squared.
Section 6 of 9
Solving Triangles
(i) Right-angled triangle
$\sin A = \dfrac{\text{opp}}{\text{hyp}}$ $\cos A = \dfrac{\text{adj}}{\text{hyp}}$ $\tan A = \dfrac{\text{opp}}{\text{adj}}$
$H^2 = O^2 + A^2$
Need 2 pieces of info to find 3rd.
(ii) Non right-angled triangle
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$
$a^2 = b^2 + c^2 - 2bc\cos A$
Area $= \tfrac{1}{2}\,ab\sin C$
Need 3 pieces of info to find 3rd.
Section 7 of 9
2-D diagrams
A TRIANGLE WITH NO LENGTH IS NO GOOD.
What have I got? $\Rightarrow$ Question.
What can I find?
A triangle has side 5 cm, 6 cm and 8 cm. Find area.
Area $= \tfrac{1}{2}\,bh$
Area $= \tfrac{1}{2}\,ab\sin C$
$a^2 = b^2 + c^2 - 2bc\cos A$
$8^2 = 5^2 + 6^2 - 2(5)(6)\cos A$
$64 = 25 + 36 - 60\cos A$
$60\cos A = -3$
$\cos A = \dfrac{-3}{60} = \dfrac{-1}{20}$
$= 92.86$
Area $= \tfrac{1}{2}\,(5)(6)\sin 92.86$
$= 14.98$ cm²
$= 15$ cm²
Section 8 of 9
Angles of elevation & depression
$A$ = angle of elevation.
$B$ = angle of depression.
Measured with a clinometer.
(i) Height of a building
A boy is 1.5 m tall. He is 6 m from base of a vertical building. He measures angle of elevation at 49°. Find height of building.
$\tan A = \dfrac{\text{opp}}{\text{adj}}$
$\tan 49 = \dfrac{h}{6}$
$6\tan 49 = h$
$6.9 = h$
Building $6.9 + 1.5 = 8.4$ m
(ii) $A = 35°,\;\; B = 62°.$ Find $h$
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$
$\dfrac{a}{\sin 35} = \dfrac{8}{\sin 27}$
$a = \dfrac{8\sin 35}{\sin 27} = 10.1$
$\sin 62 = \dfrac{h}{10.1}$
$10.1\sin 62 = h$
$8.91 = h$
$h = 8.9$
Section 9 of 9
Exam questions
(i) The ambiguous case
In a triangle $pqr$, $|\angle pqr| = 30°$, $|qr| = 15$ and $|rp| = 5\sqrt{3}$.
Find two values for $|\angle qpr|$ and sketch the two resulting triangles.
Calculate the ratio of the areas of the two triangles.
$\dfrac{\sin A}{a} = \dfrac{\sin B}{b}$
$\dfrac{\sin A}{15} = \dfrac{\sin 30}{5\sqrt{3}}$
$\sin A = \dfrac{15\sin 30}{5\sqrt{3}}$
$\sin A = \dfrac{\sqrt{3}}{2}$
$A = 60^\circ$ or $A = 120^\circ$ $(180-\theta)$
$A = 60^\circ$: Area $= \tfrac{1}{2}\,bh$ $(C=90,\ \sin 90 = 1)$ $= \tfrac{1}{2}(15)(5\sqrt{3}) = \dfrac{75\sqrt{3}}{2}$
$A = 120^\circ$: Area $= \tfrac{1}{2}(5\sqrt{3})(5\sqrt{3})\sin 120 = \dfrac{75\sqrt{3}}{4}$
$\dfrac{75\sqrt{3}}{2} : \dfrac{75\sqrt{3}}{4} = \dfrac{1}{2} : \dfrac{1}{4} = 2 : 1$
(ii) Show the area is 2.7 m²
Show that the area of the triangle $pqr$, correct to one decimal place, is 2.7 m², if $|pr| = \sqrt{8}$ m, $|\angle rpq| = 30°$ and $|\angle pqr| = 45°$.
$\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$
$\dfrac{a}{\sin 30} = \dfrac{\sqrt{8}}{\sin 45}$
$a = \dfrac{\sqrt{8}\,\sin 30}{\sin 45} = 2$
Area $= \tfrac{1}{2}\,ab\sin C$
$= \tfrac{1}{2}(2)\sqrt{8}\,\sin 105 = 2.7$ m²
SUM
The lot in one box
Non-right-angled triangle toolkit
1.Sin Rule: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$
2.Cosine Rule: $a^2 = b^2 + c^2 - 2bc\cos A$
3.Area $= \tfrac{1}{2}\,ab\sin C$
4.Opposites full? Yes $\Rightarrow$ Sin Rule. None $\Rightarrow$ Cosine Rule.
5.Largest angle is opposite the longest side.
6.Need 3 pieces of info to find the 3rd.
7.What have I got? What can I find?
End of lesson
Non-Right-Angled Triangles — HL · Mathslive.ie