Question and Answers Forum

All Questions      Topic List

UNKNOWN Questions

Previous in All Question      Next in All Question      

Previous in UNKNOWN      Next in UNKNOWN      

Question Number 90535 by 1223 last updated on 24/Apr/20

$$\mathrm{In}\mathrm{a}\mathrm{triangle}ABC,a(b\cos C-c\cos B)=$$

Answered by $@ty@m123 last updated on 24/Apr/20

$$a(b\times \frac{a^2+b^2-c^2}{2ab}-c\times \frac{c^2+a^2-b^2}{2ca})$$$$a(\frac{a^2+b^2-c^2}{2a}-\frac{c^2+a^2-b^2}{2a})$$$$\frac{1}{2}(a^2+b^2-c^2-c^2-a^2+b^2)$$$$\frac{1}{2}(2b^2-2c^2)$$$$b^2-c^2$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com