Question

Answer 1

b: \(\widehat{ABC}=180^0-24^0=156^0\)

\(\widehat{ACB}=180^0-156^0-20^0=4^0\)

Xét ΔABC có \(\dfrac{BC}{sinA}=\dfrac{AC}{sinABC}=\dfrac{AB}{sinC}\)

=>\(\dfrac{BC}{sin20}=\dfrac{8.5}{sin4}\)

=>\(BC\simeq41.68\left(m\right)\)

Xét ΔCHB vuông tại H có 

\(sin\widehat{CBH}=\dfrac{CH}{CB}\)

=>\(\dfrac{CH}{41.68}=sin24\)

=>\(CH\simeq41.68\cdot sin24\simeq16.95\left(m\right)\)

a:

\(sin^2a+cos^2a=1\)

=>\(sin^2a=1-0.16=0.84=\dfrac{21}{25}\)

=>\(sina=\dfrac{\sqrt{21}}{5}\)

\(tana=\dfrac{sina}{cosa}=\dfrac{\sqrt{21}}{5}:\dfrac{2}{5}=\dfrac{\sqrt{21}}{2}\)

\(cota=\dfrac{1}{tana}=1:\dfrac{\sqrt{21}}{2}=\dfrac{2\sqrt{21}}{21}\)