If \( \triangle A B C \) is a right triangle such that \( \angle C=90^{\circ}, \angle A=45^{\circ} \) and \( B C=7 \) units. Find \( \angle B, A B \) and \( A C \).
Given:
In a right triangle \( A B C \), right angled at \( C \), \( \angle A=45^{\circ} \) and \( B C=7 \) units.
To do:
We have to find \( \angle B, A B \) and \( A C \).
Solution:
We know that sum of the angles in a triangle is $180^o$.
Therefore,
$\angle A+\angle B+\angle C=180^o$
$\angle B+45^o+90^o=180^o$
$\angle B=180^o-135^o$
$\angle B=45^o$
$\cos\ B=\frac{BC}{AB}$
$\cos\ 45^o=\frac{BC}{AB}$
$\frac{1}{\sqrt2}$=\frac{7}{AB}$ (Since $\cos 45^{\circ}=\frac{1}{\sqrt2}$)
$AB=7\sqrt2$
$\sin\ B=\frac{AC}{AB}$
$\sin\ 45^o=\frac{AC}{AB}$
$\frac{1}{\sqrt2}$=\frac{AC}{7\sqrt2}$ (Since $\sin 45^{\circ}=\frac{1}{\sqrt2}$)
$AC=7$
The value of angle $B$ is $45^{\circ}$, the sides $AB$ and $AC$ are $7\sqrt2$units and $7$ units respectively.
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