IB Maths AA HL Sl 3 2 Triangle Trigonometry Questions

Practise choosing right-triangle, sine-rule, cosine-rule or area methods, combining them with bearings or Pythagoras and checking units and ambiguity.

Exam points

Apply sine, cosine and tangent in right-angled triangles to determine unknown sides or angles, identifying the relevant opposite, adjacent and hypotenuse sides.
Use the sine rule, cosine rule and area formula ½ab sin C to solve non-right-angled triangle problems, selecting the formula that matches the given sides and angles.

[Maximum number: 5]

Consider the triangle PQR where $\mathrm{Q} \hat{\mathrm{PR}}=30^{\circ}, \mathrm{PQ}=(x+2) \mathrm{cm}$ and $\mathrm{PR}=(5-x)^{2} \mathrm{~cm}$ , where -2<x<5.

(a)

Show that the area, $A \mathrm{~cm}^{2}$ , of the triangle is given by $A=\frac{1}{4}\left(x^{3}-8 x^{2}+5 x+50\right)$ .

[ 2 ]

use of $A=\frac{1}{2} q r \sin \theta$ to obtain $A=\frac{1}{2}(x+2)(5-x)^{2} \sin 30^{\circ}$ $=\frac{1}{4}(x+2)\left(25-10 x+x^{2}\right)$ $A=\frac{1}{4}\left(x^{3}-8 x^{2}+5 x+50\right)$

(b)

Find QR when the area of triangle PQR is a maximum.

[ 3 ]

$\quad \mathrm{PQ}=\frac{7}{3}$ (cm) and $\mathrm{PR}=\left(\frac{14}{3}\right)^{2}$ (cm)

\mathrm{QR}^{2}=\left(\frac{7}{3}\right)^{2}+\left(\frac{14}{3}\right)^{4}-2\left(\frac{7}{3}\right)\left(\frac{14}{3}\right)^{2} \cos 30^{\circ}

$=391.702 \ldots$
QR $=19.8(\mathrm{~cm})$

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IB Maths AA HL Sl 3 2 Triangle Trigonometry Questions question 1

Question (a)

Question (b)