Question
If $\triangle\text{ABC}$ is a right triangle such that $\angle\text{C} = 90^\circ,\angle\text{A}=45^\circ$ and BC = 7 units. Find $\angle\text{B}$ AB and AC.
✓
Answer
We have $\angle\text{C}=90^\circ,\angle\text{A}=45^\circ$ and $\text{BC}=7\text{ units}$ In $\triangle\text{ABC},$ $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ $\Rightarrow45^\circ+\angle\text{B}+90^\circ=180^\circ$ $\Rightarrow\angle\text{B}=180^\circ-135^\circ$ $\Rightarrow\angle\text{B}=45^\circ$
Now, $\tan\text{B}=\frac{\text{AC}}{\text{BC}}$ $\Rightarrow\tan45^\circ=\frac{\text{AC}}{7}$ $\Rightarrow\text{AC}=\frac{\text{BC}}{\tan45^\circ}$ $\Rightarrow\text{AC}=\frac{7}{1}$ $\Rightarrow\text{AC}=7\text{ units}$ And, $\cos\text{B}=\frac{\text{BC}}{\text{AB}}$ $\Rightarrow\text{AB}=\frac{\text{BC}}{\cos45^\circ}$ $\Rightarrow\text{AB}=\frac{7}{\frac{1}{\sqrt{2}}}$ $\Rightarrow\text{AB}=7\sqrt{2}\text{ units}$
Now, $\tan\text{B}=\frac{\text{AC}}{\text{BC}}$ $\Rightarrow\tan45^\circ=\frac{\text{AC}}{7}$ $\Rightarrow\text{AC}=\frac{\text{BC}}{\tan45^\circ}$ $\Rightarrow\text{AC}=\frac{7}{1}$ $\Rightarrow\text{AC}=7\text{ units}$ And, $\cos\text{B}=\frac{\text{BC}}{\text{AB}}$ $\Rightarrow\text{AB}=\frac{\text{BC}}{\cos45^\circ}$ $\Rightarrow\text{AB}=\frac{7}{\frac{1}{\sqrt{2}}}$ $\Rightarrow\text{AB}=7\sqrt{2}\text{ units}$
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