Math Question and Answers


Question and Answers Forum

All Questions Topic List
Trigonometry Questions
Previous in All Question Next in All Question
Previous in Trigonometry Next in Trigonometry
Question Number 204628 by MM42 last updated on 23/Feb/24


	Answered by MM42 last updated on 23/Feb/24

	Answered by A5T last updated on 23/Feb/24

	Commented by A5T last updated on 23/Feb/24

	$\frac{s i n 60}{E B} = \frac{s i n 40}{B C} \Rightarrow E B = \frac{B C s i n 60}{s i n 40} = \frac{B C \sqrt{3}}{2 s i n 40}$ $\Rightarrow {E C}^{2} = \frac{3 {B C}^{2}}{4 {s i n}^{2} 40} + {B C}^{2} - \frac{{B C}^{2} c o s 80 \sqrt{3}}{s i n 40}$ $\Rightarrow E C = B C \sqrt{\frac{3}{4 {s i n}^{2} 40} - \frac{\sqrt{3} c o s 80}{s i n 40} + 1}$ $\frac{D C}{s i n 70} = \frac{B C}{s i n 30} \Rightarrow D C = 2 B C s i n 70$ $\frac{s i n (30 + a)}{E C} = \frac{s i n (130 - a)}{D C}$ $\Rightarrow \frac{s i n (30 ° + a)}{\sqrt{1 + \frac{3}{4 {s i n}^{2} 40 °} - \frac{\sqrt{3} c o s 80}{s i n 40 °}}} = \frac{s i n (130 ° - a)}{2 s i n 70 °}$ $\Rightarrow a = 20 °$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum