Question

Help again...its for math offering 15 points

Answer 1

Answer:

$\tan \dfrac{\theta}2 = \dfrac{1}2$

Step by step Explanation:

$\sin \theta = \dfrac{\text{Perpendicular} }{\text{Hypotenuse}} = \dfrac{12}{15}\\\\\\\cos \theta = \dfrac{\text{Base}}{\text{Hypotenuse}}= \dfrac{9}{15}\\\\\text{Now,}\\\\\tan \dfrac{\theta}2 = \dfrac{\sin \tfrac{\theta}2}{\cos \tfrac{\theta}2}\\\\\\~~~~~~~~=\dfrac{2\cos \tfrac{\theta}2 \sin \tfrac{\theta}2 }{2\cos^2 \tfrac{\theta}2}~~~~~~;\left[\text{Multiply by}~ 2\cos\tfrac{\theta}2 \right]$

         $=\dfrac{\sin \theta}{1+ \cos \theta}~~~~~~~~~~~~;[2 \sin x \cos x = \sin 2x ~ \text{and}~ 2\cos^2 x =1+\cos 2x]\\\\\\=\dfrac{\tfrac{12}{15}}{1+ \tfrac{9}{15}}\\\\\\=\dfrac{\tfrac{12}{15}}{\tfrac{24}{15}}\\\\\\=\dfrac{12}{15}\times \dfrac{15}{24}\\\\\\=\dfrac{12}{24}\\\\\\=\dfrac{1}2$