Answer:
a = √325 units, m∠B = 34.63° and m∠C=55.37°
Step-by-step explanation:
Consider in triangle ABC,
AC = b = 10 units,
AB = c = 15 units,
m∠A = 90°,
BC = a
Using law of cosine,
[tex]a^2 = b^2 + c^2 - 2bc \cos A[/tex]
[tex]a^2 = 10^2 + 15^2 - 2\times 10\times 15 \cos 90^{\circ}[/tex]
[tex]a^2 = 100 + 225[/tex]
[tex]a^2 = 325[/tex]
[tex]a=\sqrt{325}[/tex] ( negative value is not recommended because side can't be negative )
Now, Again using law of cosine,
[tex]b^2 = a^2 + c^2 - 2ac\cos B[/tex]
[tex]10^2 = 325 + 15^2 - 2\sqrt{325}(15) \cos B[/tex]
[tex]100 = 325 + 225 - 30\sqrt{325}\cos B[/tex]
[tex]100 = 550 - 30\sqrt{325}\cos B[/tex]
[tex]30\sqrt{325}\cos B=550-100[/tex]
[tex]30\sqrt{325}\cos B=450[/tex]
[tex]\cos B=\frac{450}{30\sqrt{325}}\implies m\angle B = 33.69^{\circ}[/tex]
Sum of measures of all interior angles of a triangle is supplementary,
⇒m∠A + m∠B + m∠C = 180°
⇒ 90° + 33.69° + m∠C = 180°
⇒ 123.69° + m∠C = 180°
⇒ m∠C = 180° - 123.69° = 56.31°
