Can you help me this please!!!And after that, can you do the length of side a and b please!!!

Question

madge20 · Accepted Answer

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

\(\begin{gathered} \angle C=90^{\circ} \\ \angle A=53^{\circ}16^{\prime} \\ AB=961\text{ m} \end{gathered}\)

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

\(\begin{gathered} \angle B=180-90-53^{\circ}16^{\prime} \\ \angle B=90-53^{\circ}16^{\prime} \\ \angle B=89^{\circ}60^{\prime}-53^{\circ}16^{\prime} \\ \angle B=36^{\circ}44^{\prime} \end{gathered}\)

To find a and b, we will use the trigonometry ratios sine and cosine

\(\begin{gathered} sinA=\frac{opposite}{hypotenuse} \\ sinA=\frac{a}{AB} \\ sin53^{\circ}16^{\prime}=\frac{a}{961} \end{gathered}\)

By using the cross multiplication

\(\begin{gathered} a=961\times sin53^{\circ}16^{\prime} \\ a=770.1721392 \\ a=770\text{ m} \end{gathered}\)\(\begin{gathered} cosA=\frac{adjacent}{hypotenuse} \\ cosA=\frac{b}{AB} \\ cos53^{\circ}16^{\prime}=\frac{b}{961} \end{gathered}\)

By using the cross multiplication

\(\begin{gathered} b=961\times cos53^{\circ}16^{\prime} \\ b=574.7659315 \\ b=575\text{ m} \end{gathered}\)