Question

Cho hình thang cân ABCD có đáy nhỏ AB = AD. Biết \(\tan\widehat{BDC}=\dfrac{3}{4}\). Tính các giá trị lượng giác của \(\widehat{BAD}\)

Answer 1

Ta có :

\(\widehat{ABD}=\widehat{ADB}\)

\(\widehat{ABD}=\widehat{BDC}\)

\(\Rightarrow\widehat{BDC}=\widehat{ADB}\)

Suy ra \(\widehat{BAD}=\pi-2\widehat{BDC}\)

Từ đó ta có :

\(\tan\widehat{BAD}=-\tan2\widehat{BDC}=-\dfrac{2\tan\widehat{BDC}}{1-\tan^2\widehat{BDC}}=-\dfrac{2.\dfrac{3}{4}}{1-9\cdot16}=-\dfrac{3}{2}.\dfrac{16}{7}=-\dfrac{24}{7}\)Vì \(\dfrac{\pi}{2}< \widehat{BAD}< \pi\) nên \(\cos\widehat{BAD}< 0\)
Do đó : \(\cos\widehat{BAD}=-\dfrac{1}{\sqrt{1+\tan^2\widehat{BAD}}}=-\dfrac{1}{\sqrt{1+\dfrac{576}{49}}}=-\dfrac{7}{25}\)

\(\sin\widehat{BAD}=\cos\widehat{BAD}\tan\widehat{BAD}=\dfrac{-7}{25}.\dfrac{-24}{7}=\dfrac{24}{25}\)