Question

One of the exterior angles of a triangle $70°$and the interior opposite angles are in the ratio $3:4$ Find the angles of the triangle.

Answer 1

Given, the exterior angle $=70°$

and the ratio of interior angles $3:4$

We have to find the value of all interior angles.

Let $\angle 1,\angle 2and\angle 3$ are interior angles and $\angle 4$is the exterior angle of the $△ABC$

Let $\angle 1=3xand\angle 2=4x(\sin ce\angle 1:\angle 2=3:4)$

We know that from the exterior angle property of a triangle

the sum of opposite interior angles$=$ exterior angles

$$\begin{gathered} \angle 1+\angle 2=\angle 4 \\ \Rightarrow 3x+4x=70°(\sin ce\angle 4=70°) \\ \Rightarrow 7x=70° \\ \Rightarrow x=\frac{70°}{7}=10° \end{gathered}$$

then ,$$\begin{gathered} \angle 1=3x=3\times 10°=30° \\ and\angle 2=4x=4\times 10°=40° \end{gathered}$$

Now, we have to find the third angle $\angle 3$

By the angle sum property of a triangle,

$$\begin{gathered} \angle 1+\angle 2+\angle 3=180° \\ \Rightarrow 30°+40°+\angle 3=180° \\ \Rightarrow 70°+\angle 3=180° \\ \Rightarrow \angle 3=180°-70°=110° \end{gathered}$$

Hence, the angles are $\angle 1=30°,\angle 2=40°and\angle 3=110°$