Question

find the interior angles of the triangle whose vertices are A(4,3)B(-2,2)and C(2,-8)​

Answer 1

Given that,

The triangle whose vertices are,

A= 4,3

B= -2,2

C= 2,-8

We need to calculate the distance between A and B

Using formula of distance

$AB=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Put the value into the formula

$AB=\sqrt{(4-(-2))^2+(3-2)^2}$

$AB=\sqrt{37}$

We need to calculate the distance between B and C

Using formula of distance

$BC=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Put the value into the formula

$BC=\sqrt{(-2-2)^2+(2-(-8))^2}$

$BC=\sqrt{116}$

We need to calculate the distance between C and A

Using formula of distance

$CA=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

Put the value into the formula

$CA=\sqrt{(2-4)^2+(-8-3)^2}$

$CA=\sqrt{125}$

We need to calculate the angle C

Using law of cosine

$AB^2=BC^2+AC^2-2BC\times AC\cos C$

Put the value into the formula

$(\sqrt{37})^2=(\sqrt{116})^2+(\sqrt{125})^2-2\times\sqrt{116}\times\sqrt{125}\cos C$

$-240.83\cos C=37-116-125$

$C=\cos^{-1}(\dfrac{204}{240.83})$

$C=32.10^{\circ}$

We need to calculate the angle B

Using law of sines

$\dfrac{AB}{\sin C}=\dfrac{AC}{\sin B}$

Put the value into the formula

$\dfrac{\sqrt{37}}{\sin(32.10)}=\dfrac{\sqrt{125}}{\sin B}$

$B=\sin^{-1}(\dfrac{\sqrt{125}\times\sin(32.10)}{\sqrt{37}})$

$B=77.1^{\circ}$

We need to calculate the angle of A

Using formula for angle A

$A=180-77.1-32.10$

$A=70.8^{\circ}$

Hence, The interior angles of the triangle are 32.10°, 77.1° and 70.8°