a, Xét ΔAHB, có:
${\mathrm{AB}}^2$ = ${\mathrm{AH}}^2$ + ${\mathrm{HB}}^2$
hay ${30}^2$​ = ${24}^2$ + ${\mathrm{HB}}^2$
=> ${\mathrm{HB}}^2$​ = 900 - 576
=> ${\mathrm{HB}}^2$ = 324
=> HB = 18cm
Áp dụng hệ thức về cạnh và đường cao trong tam giác, ta có:
${\mathrm{AH}}^2$ = HB.HC
=> ${24}^2$​ = 18.HC
=> HC = 32cm
BC = HB + HC = 32 + 18 = 50cm
Xét ΔBAC vuông tại A, có:
${\mathrm{AC}}^2$ + ${\mathrm{AB}}^2$ = ${\mathrm{BC}}^2$
=> ${50}^2$​ = ${\mathrm{AC}}^2$ + ${30}^2$
=> AC = 40cm
sinB = $\frac{\mathrm{đ}}{\mathrm{h}}$​ = $\frac{\mathrm{AC}}{\mathrm{BC}}$ = $\frac{40}{50}$ = $\frac{4}{5}$
cosB = $\frac{\mathrm{k}}{\mathrm{h}}$ = $\frac{\mathrm{AB}}{\mathrm{BC}}$ = $\frac{30}{50}$ = $\frac{3}{5}$

tanB = $\frac{\mathrm{đ}}{\mathrm{k}}$ = $\frac{\mathrm{AC}}{\mathrm{AB}}$ = $\frac{40}{30}$ = $\frac{4}{3}$

cotB = $\frac{\mathrm{k}}{\mathrm{đ}}$ = $\frac{\mathrm{AB}}{\mathrm{AC}}$ = $\frac{30}{40}$ = $\frac{3}{4}$
Vì $\hat{\mathrm{B}}$ = $\hat{\mathrm{C}}$ => sinB = sinC, tanB = cotC, cotB = tanC, cosB = cosC

b, Xét ΔAHB, có:
${\mathrm{AB}}^2$ = ${\mathrm{AH}}^2$ + ${\mathrm{HB}}^2$
hay ${\mathrm{AB}}^2$​ = $2^2$ + ${\left(2\sqrt{3}\right)}^2$
=> ${\mathrm{AB}}^2$​ = 16 => AB = 4cm
Áp dụng HTVCVĐCTΔV, ta có:
${\mathrm{AH}}^2$ = HB.HC
hay ${\left(2\sqrt{3}\right)}^2$ = 2.HC
=> HC = 6cm
BC = HB + HC = 6 + 2 = 8cm
Xét ΔABC, có:
${\mathrm{BC}}^2$ = ${\mathrm{AB}}^2$ + ${\mathrm{AC}}^2$
hay $8^2$ = $4^2$ + ${\mathrm{AC}}^2$
=> ${\mathrm{AC}}^2$​ = 48
=> AC = $4\sqrt{3}$
sinB = $\frac{\mathrm{đ}}{\mathrm{h}}$ = $\frac{\mathrm{AB}}{\mathrm{AC}}$ = $\frac{4\sqrt{3}}{8}$ = $\frac{\sqrt{3}}{2}$
cosB = $\frac{\mathrm{k}}{\mathrm{h}}$ = $\frac{AB}{BC} = \frac{2}{8} = \frac{1}{4}$

tanB = $\frac{\mathrm{đ}}{\mathrm{k}}=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{4\sqrt{3}}{2}=2\sqrt{2}$

cotB = $\frac{k}{đ}=\frac{AB}{AC}=\frac{2}{4\sqrt{3}}=\frac{\sqrt{3}}{6}$
Vì $\hat{\mathrm{B}}$ = $\hat{\mathrm{C}}$ => sinB = sinC, tanB = cotC, cotB = tanC, cosB = cosC

c, BC = BH + CH = 25 + 9 = 34cm
Áp dụng HTVCVĐCΔV, ta có:
${\mathrm{AH}}^2 = \mathrm{HB}.\mathrm{HC}$
$$\begin{gathered} {\mathrm{AH}}^2= 25.9 \\ => {\mathrm{AH}}^2 = 225 \\ => \mathrm{AH} = 15\mathrm{cm} \end{gathered}$$​
 Xét ΔBHA, có:
$$\begin{gathered} {\mathrm{AB}}^2 = {\mathrm{AH}}^2 + {\mathrm{HB}}^2 \\ => {\mathrm{AB}}^2 = {15}^2 + {25}^2 \\ => {\mathrm{AB}}^2 = 850 \\ => \mathrm{AB} = 5\sqrt{34 \mathrm{cm}} \end{gathered}$$

sinB = $\frac{\mathrm{đ}}{\mathrm{h} } = \frac{\mathrm{AB}}{\mathrm{BC}} = \frac{3\sqrt{34}}{34}$

cosB = $\frac{\mathrm{k}}{\mathrm{h}}= \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{5\sqrt{34}}{34}$

tanB = $\frac{\mathrm{đ}}{\mathrm{h}}=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{3\sqrt{34}}{5\sqrt{34}}$

cotB = $\frac{\mathrm{k}}{\mathrm{đ}}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{5\sqrt{34}}{3\sqrt{34}}$
Vì $\hat{\mathrm{B}}$ = $\hat{\mathrm{C}}$ => sinB = sinC, tanB = cotC, cotB = tanC, cosB = cosC