Question

B=10, c=15, a=.90 degrees solve the triangle using the law

ofcosines.
b=10, c=15, a=90 degrees solve the trianlge using the law
ofcosines.

Answer 1

a = √325 units, m∠B = 34.63° and m∠C=55.37°

Explanation:

Consider in triangle ABC,

AC = b = 10 units,

AB = c = 15 units,

m∠A = 90°,

BC = a

Using law of cosine,

`a^2 = b^2 + c^2 - 2bc \cos A`

`a^2 = 10^2 + 15^2 - 2* 10* 15 \cos 90^(\circ)`

`a^2 = 100 + 225`

`a^2 = 325`

`a=√(325)` ( negative value is not recommended because side can't be negative )

Now, Again using law of cosine,

`b^2 = a^2 + c^2 - 2ac\cos B`

`10^2 = 325 + 15^2 - 2√(325)(15) \cos B`

`100 = 325 + 225 - 30√(325)\cos B`

`100 = 550 - 30√(325)\cos B`

`30√(325)\cos B=550-100`

`30√(325)\cos B=450`

`\cos B=(450)/(30√(325))\implies m\angle B = 33.69^(\circ)`

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°