Question

In $△\mathrm{ABC}$, right-angled at $B$,$\mathrm{AB}=24\mathrm{cm},\mathrm{BC}=7\mathrm{cm}$. Determine: $\sin \left(C\right),\cos \left(C\right)$

Answer 1

Step- 1: Find hypotenuse of the given triangle:

Given: $△\mathrm{ABC}$ right-angled at $B$, $\mathrm{AB}=24\mathrm{cm},\mathrm{BC}=7\mathrm{cm}$$\Rightarrow AC$ is the hypotenuse.

According to Pythagoras theorem, if $a,b,c$ are sides of the triangle where $b>a,c$ then $b=\sqrt{a^2+c^2}$

$$\begin{gathered} \mathrm{AB}=24\mathrm{cm},\mathrm{BC}=7\mathrm{cm} \\ \therefore \mathrm{AC}=\sqrt{{\mathrm{AB}}^2+{\mathrm{BC}}^2}\mathbf{} \\ =\sqrt{{24}^2+7^2} \\ =\sqrt{625} \\ \Rightarrow \mathrm{AC}=25\mathrm{cm} \end{gathered}$$

Step- 2: Find the required values:

We know that for a right angled triangle $\sin \left(\theta \right)=\frac{P}{H},\cos \left(\theta \right)=\frac{B}{H}$ where $P,B,H$ are perpendicular, base and hypotenuse respectively.

$$\begin{gathered} \sin \left(C\right)=\frac{\mathrm{P}}{\mathrm{H}}=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{24}{25} \\ \cos \left(C\right)=\frac{\mathrm{B}}{\mathrm{H}}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{7}{25} \end{gathered}$$

Hence the required values of $\sin \left(C\right),\cos \left(C\right)$ are $\frac{24}{25},\frac{7}{25}$ respectively.