Question

In quadrilateral ABQP, if angle PAB = 60, angle QAB = 45, angle PBA = 30, angle QBA = 75 and AB=20√3, then find the measure of PQ.

Note: Could you please use trigonometry?

Answer 1

Answer to Question #51111 - Math - Trigonometry

Question

In quadrilateral ABQP, if angle PAB = 60, angle QAB = 45, angle PBA = 30, angle QBA = 75 and AB=20√3, then find the measure of PQ.

Solution

Let's imagine two triangles ABP and ABQ

In triangle ABP we have $\angle APB = 180{}^{\circ} - \angle PAB - \angle PBA = 180{}^{\circ} - 60{}^{\circ} - 30{}^{\circ} = 90{}^{\circ}$, hence

$$AP = AB \cdot \cos 60{}^{\circ} = 20\sqrt{3} \cdot \frac{1}{2} = 10\sqrt{3}; BP = AB \cdot \sin 60{}^{\circ} = 20\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 30.$$

In triangle ABQ we have $\angle AQB = 180{}^{\circ} - \angle QAB - \angle QBA = 180{}^{\circ} - 45{}^{\circ} - 75{}^{\circ} = 60{}^{\circ}$.

Using law of sines, $\frac{AB}{\sin \angle AQB} = \frac{BQ}{\sin \angle QAB}$, hence

$$BQ = \frac{\sin \angle QAB}{\sin \angle AQB} AB = \frac{\sin 45{}^{\circ}}{\sin 60{}^{\circ}} AB = \frac{1}{\sqrt{2}} \cdot \frac{2}{\sqrt{3}} \cdot 20\sqrt{3} = 20\sqrt{2}.$$

Angle $\angle PBQ = \angle QBA - \angle PBA = 75{}^{\circ} - 30{}^{\circ} = 45{}^{\circ}$.

Using law of cosines,

$$\begin{array}{l} PQ^2 = BP^2 + BQ^2 - 2BP \cdot BQ \cdot \cos \angle PBQ = \\ = 30^2 + (20\sqrt{2})^2 - 2 \cdot 30 \cdot 20\sqrt{2} \cdot \cos 45{}^{\circ} = 500, \text{ hence } PQ = \sqrt{500} \approx 22.4 \end{array}$$

Answer: 22.4.

www.AssignmentExpert.com