To solve a right triangle means to find all unknown angles and sides
From the given figure we can seeL
A right triangle ABC, where
[tex]\begin{gathered} \angle C=90^{\circ} \\ \angle A=53^{\circ}16^{\prime} \\ AB=961\text{ m} \end{gathered}[/tex]Since the sum of angles of a triangle is 180 degrees, then we can find angle B by subtraction angles C and A from 180
[tex]\begin{gathered} \angle B=180-90-53^{\circ}16^{\prime} \\ \angle B=90-53^{\circ}16^{\prime} \\ \angle B=89^{\circ}60^{\prime}-53^{\circ}16^{\prime} \\ \angle B=36^{\circ}44^{\prime} \end{gathered}[/tex]To find a and b, we will use the trigonometry ratios sine and cosine
[tex]\begin{gathered} sinA=\frac{opposite}{hypotenuse} \\ sinA=\frac{a}{AB} \\ sin53^{\circ}16^{\prime}=\frac{a}{961} \end{gathered}[/tex]By using the cross multiplication
[tex]\begin{gathered} a=961\times sin53^{\circ}16^{\prime} \\ a=770.1721392 \\ a=770\text{ m} \end{gathered}[/tex][tex]\begin{gathered} cosA=\frac{adjacent}{hypotenuse} \\ cosA=\frac{b}{AB} \\ cos53^{\circ}16^{\prime}=\frac{b}{961} \end{gathered}[/tex]By using the cross multiplication
[tex]\begin{gathered} b=961\times cos53^{\circ}16^{\prime} \\ b=574.7659315 \\ b=575\text{ m} \end{gathered}[/tex]