Answer:
• BC = 7√5 units.
,
• m∠B=41.8°
,
• m∠C=48.2°
Explanation:
First, find the length of the third side, BC using the Pythagorean theorem.
[tex]\begin{gathered} AB^2=AC^2+BC^2 \\ 21^2=14^2+BC^2 \\ BC^2=21^2-14^2 \\ BC=\sqrt{21^2-14^2} \\ BC=\sqrt{245} \\ BC=\sqrt{49\times5} \\ BC=7\sqrt{5} \end{gathered}[/tex]
The length of the third side, BC is 7√5 units.
(b)Next, find the measure of angle B.
• The side length ,opposite, B = 14
,
• The length of the ,hypotenuse, = 21
From trigonometric ratios:
[tex]\begin{gathered} \sin B=\frac{Opposite}{Hypotenuse} \\ \sin B=\frac{14}{21} \\ T\text{ake the arcsine of both sides:} \\ B=\arcsin(\frac{14}{21}) \\ B\approx41.8\degree \end{gathered}[/tex]
The measure of angle B is 41.8 degrees.
(c)Finally, find the measure of angle C.
In a right triangle, the sum of the two acute angles is 90 degrees, therefore:
[tex]\begin{gathered} m\angle B+m\angle C=90\degree \\ 41.8\operatorname{\degree}+m\operatorname{\angle}C=90\operatorname{\degree} \\ m\operatorname{\angle}C=90\operatorname{\degree}-41.8\operatorname{\degree} \\ m\operatorname{\angle}C=48.2\operatorname{\degree} \end{gathered}[/tex]
The measure of angle C is 48.2 degrees.
