Question

Im giving 100 Points to anyone who could solve this for me please.

Answer 1

Answer:

AB = 27.2

BC = 33.5

AC = 50.4

∠A = 38°

∠ABC = 112°

Step-by-step explanation:

Trigonometric ratios

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

• $\theta$ is the angle
• O is the side opposite the angle
• A is the side adjacent the angle
• H is the hypotenuse (the side opposite the right angle)

$\implies \sf \cos(30^{\circ})=\dfrac{29}{BC}$

$\implies \sf BC=\dfrac{29}{\cos(30^{\circ})}$

$\implies \sf BC=\dfrac{58\sqrt{3}}{3}=33.5\:(nearest\:tenth)$

$\implies \sf \tan(30^{\circ})=\dfrac{BD}{29}$

$\implies \sf BD=29\tan(30^{\circ})$

$\implies \sf BD=\dfrac{29\sqrt{3}}{3}$

$\implies \sf \sin(38^{\circ})=\dfrac{BD}{AB}$

$\implies \sf AB=\dfrac{\dfrac{29\sqrt{3}}{3}}{\sin(38^{\circ})}$

$\implies \sf AB=27.2\:(nearest\:tenth)$

$\implies \sf \tan(38^{\circ})=\dfrac{BD}{AD}$

$\implies \sf AD=\dfrac{\dfrac{29\sqrt{3}}{3}}{\tan(38^{\circ})}$

$\implies \sf AD=21.4\:(nearest\:tenth)$

$\implies \sf AC=AD+DC=21.4+29=50.4$

The interior angles of a triangle sum to 180°

⇒ ∠A + 52° + 90° = 180°

⇒ ∠A = 180° - 90° - 52°

⇒ ∠A = 38°

⇒ ∠ABC + 38° + 30° = 180°

⇒ ∠ABC = 180° - 38° - 30°

⇒ ∠ABC = 112°

**I have checked the measures using a graphing programme - see attached**