Giải thích các bước giải:

Bài 20:

1.Ta có $\Delta ABC$ vuông tại $A\to AC=\sqrt{BC^2-AB^2}=24$

$\to \sin B=\dfrac{AC}{BC}=\dfrac{12}{13}$

      $\cos B=\dfrac{AB}{BC}=\dfrac{5}{13}$

      $\tan B=\dfrac{AC}{AB}=\dfrac{12}5$

      $\cot B=\dfrac{AB}{AC}=\dfrac5{12}$

2.Ta có: $\Delta ABC$ vuông tại $A$

$\to AC=\sqrt{BC^2-AB^2}=4\sqrt7$

$\to \sin B=\dfrac{AC}{BC}=\dfrac{\sqrt7}{4}$

      $\cos B=\dfrac{AB}{BC}=\dfrac{3}{4}$

      $\tan B=\dfrac{AC}{AB}=\dfrac{\sqrt7}3$

      $\cot B=\dfrac{AB}{AC}=\dfrac3{\sqrt7}$

Bài 21:

1.Ta có $\Delta ABC$ vuông tại $A$

$\to BC=\sqrt{AB^2+AC^2}=75$

Mà $AH\perp BC$

$\to AH\cdot BC=AB\cdot AC$

$\to AH=\dfrac{504}{25}$

$\to BH=\sqrt{AB^2-AH^2}=\dfrac{147}{25}$

$\to \sin\widehat{BAH}=\dfrac{BH}{AB}=\dfrac7{25}$

      $\cos\widehat{BAH}=\dfrac{AH}{AB}=\dfrac{24}{25}$

      $\tan\widehat{BAH}=\dfrac{BH}{AH}=\dfrac{7}{24}$

      $\cot\widehat{BAH}=\dfrac{AH}{BH}=\dfrac{24}7$

$\to \widehat{BAH}=\arcsin(\dfrac7{25})\approx 16.26^o$

b.Ta có:

$\widehat{CAH}=90^o-\widehat{BAH}$

$\to \sin\widehat{CAH}=\cos\widehat{BAH}=\dfrac{24}{25}$

      $\cos\widehat{CAH}=\sin\widehat{BAH}=\dfrac7{25}$

      $\tan\widehat{CAH}=\cot\widehat{BAH}=\dfrac{24}7$

      $\cot\widehat{CAH}=\tan\widehat{BAH}=\dfrac7{24}$

Bài 22:

Đổi $3cm=30mm$

1.Ta có $\Delta ABC$ vuông tại $A$

$\to BC=\sqrt{AB^2+AC^2}=34$

$\to \sin B=\dfrac{AC}{BC}=\dfrac{15}{17}$

     $\cos B=\dfrac{AB}{BC}=\dfrac6{17}$

     $\tan B=\dfrac{AC}{AB}=\dfrac{15}{6}$

     $\cot B=\dfrac{AB}{AC}=\dfrac6{15}$

     $\sin C=\dfrac{AB}{BC}=\dfrac6{17}$

     $\cos C=\dfrac{AC}{BC}=\dfrac{15}{17}$

     $\tan C=\dfrac{AB}{AC}=\dfrac6{15}$

     $\cot C=\dfrac{AC}{AB}=\dfrac{15}6$

$\to \hat B=\arcsin (\dfrac{15}{17})=61.93^o$

$\to \hat C=90^o-\hat B\approx 28.07^o$