In a right angled triangle $\Delta$ ABC, $\angle$ $B = 90^\circ$ and $\sin$ $A = 5/\sqrt{29}.$ Then which of the following is not true?

The correct answer is (D) $\sec$ C = $\sqrt{29}$/2.

• In right-angled triangle $\mathrm{\Delta }ABC$$\Delta$ ABC with $\mathrm{\angle }B={90}^{\circ }$$\angle$ B = 90^$\circ$:
• Given $\sin A=\frac{\text{Opposite}}{\text{Hypotenuse}}=\frac{BC}{AC}=\frac{5}{\sqrt{29}}$$\sin$ A = $\frac{\text{Opposite}}{\text{Hypotenuse}}$ = $\frac{BC}{AC}$ = $\frac{5}{\sqrt{29}}$.
• By Pythagoras theorem: $AB =$$\sqrt{AC^2 - BC^2}$ = $\sqrt{29 - 25}$ = $\sqrt{4}$ = 2.
• Now evaluating trigonometric ratios for angle $C$ (where opposite side is $AB = 2$ and adjacent side is $BC = 5$):
• $\cos A=\frac{AB}{AC}=\frac{2}{\sqrt{29}}\text{ (True)},\tan C=\frac{AB}{BC}=\frac{2}{5}\text{ (True)},\cos C=\frac{BC}{AC}=\frac{5}{\sqrt{29}}\text{ (True)}$$\cos$ A = $\frac{AB}{AC}$ = $\frac{2}{\sqrt{29}}$ $\text{ (True)}$, \quad $\tan$ C = $\frac{AB}{BC}$ = $\frac{2}{5}$ $\text{ (True)}$, \quad $\cos$ C = $\frac{BC}{AC}$ = $\frac{5}{\sqrt{29}}$ $\text{ (True)}$
• $\sec C=\frac{1}{\cos C}=\frac{AC}{BC}=\frac{\sqrt{29}}{5}\mathrm{\ne }\frac{\sqrt{29}}{2}$$\sec$ C = $\frac{1}{\cos C}$ = $\frac{AC}{BC}$ = $\frac{\sqrt{29}}{5}$ $\ne$ $\frac{\sqrt{29}}{2}$
• Therefore, the statement $\sec C=\sqrt{29}\mathrm{/}2$$\sec$ C = $\sqrt{29}$/2 is not true, making (D) the correct choice.