Question

Name the type of following triangle.
$\mathrm{\Delta }ABC$ with AB = 8.7cm, AC = 7cm and BC = 6cm.

Answer 1

Hint: Triangles are classified in three types on the basis of length of its sides and in three types on the basis of its angles. Lengths of sides are given in question and we can calculate its angles by using sine or cosine formula.

Complete step-by-step solution -
For any $\mathrm{\Delta }ABC$ :
Sine formula: ${\displaystyle \frac{a}{\sin A}}={\displaystyle \frac{b}{\sin B}}={\displaystyle \frac{c}{\sin C}}$

Cosine formula:
$\begin{array}{rl}& \cos A={\displaystyle \frac{b^2+c^2-a^2}{2bc}}\\ & \cos B={\displaystyle \frac{a^2+c^2-b^2}{2ac}}\\ & \cos C={\displaystyle \frac{a^2+b^2-c^2}{2ab}}\end{array}$

Given triangle,

On the basis of length of sides, triangles are of three types:
1) Equilateral triangle – A triangle whose all three sides are of equal length.
2) Isosceles triangle – A triangle whose two sides are of equal length and third one is of different length.
3) Scalene triangle – A triangle whose all sides are of different length.
Since, the given triangle has all sides of different lengths, it is a scalene triangle.
On the basis of angles, triangles are of three types:
1) Acute – angled triangle – A triangle whose all angles are acute angle. (i.e. $angle<90{}^{\circ }$)
2) Right – angle triangle – A triangle whose one angle is equal to $90{}^{\circ }$. (i.e. right angle)
3) Obtuse – angled triangle – A triangle whose one angle is obtuse. (i.e. $angle>90{}^{\circ }$)
By using cosine formula –
$\begin{array}{rl}& \cos A={\displaystyle \frac{b^2+c^2-a^2}{2bc}}\\ & \Rightarrow \cos A={\displaystyle \frac{{\left(7\right)}^2+{\left(8.7\right)}^2-{\left(6\right)}^2}{2\left(7\right)\left(8.7\right)}}\\ & \Rightarrow \cos A={\displaystyle \frac{(49+75.69)-36}{121.8}}\\ & \Rightarrow \cos A={\displaystyle \frac{124.69}{121.8}}\\ & \Rightarrow \cos A=0.73\end{array}$
Similarly,
$\begin{array}{rl}& \cos B={\displaystyle \frac{a^2+c^2-b^2}{2ac}}\\ & \Rightarrow \cos B={\displaystyle \frac{{\left(6\right)}^2+{\left(8.7\right)}^2-{\left(7\right)}^2}{2\left(6\right)\left(8.7\right)}}\\ & \Rightarrow \cos B={\displaystyle \frac{62.69}{104.4}}\\ & \Rightarrow \cos B=0.60\end{array}$
Similarly,
$$\begin{array}{rl}& \cos C={\displaystyle \frac{a^2+b^2-c^2}{2ab}}\\ & \Rightarrow \cos C={\displaystyle \frac{{\left(6\right)}^2+{\left(7\right)}^2-{\left(8.7\right)}^2}{2\left(6\right)\left(7\right)}}\\ & \Rightarrow \cos C={\displaystyle \frac{9.31}{84}}\\ & \Rightarrow \cos C=0.11\end{array}$$
Graph of cos function:

From the graph, we can see that for acute angles, $\cos \theta$ is positive and for obtuse angles $\cos \theta$ is negative.
Since, cos A, cos B and cos C are positive, A, B and C are acute angles.
So, the given triangle is acute – angles triangle.
Hence, the given triangle is an acute and scalene triangle.

Note: The possibility for the mistake that you put the length of sides wrong in the formula. Be careful that
a = BC
b = AC
c = AB