Important Solutions of Triangle Formulas - Practice Questions & MCQ

Problems in which all three sides are given

Till now we have learned different rules and properties of triangles. Now we will see their application in different situations.

In a triangle, there are six variables, three sides (say a, b, c) and three angles (say A, B, C). If any three of these six variables (except all the angles A, B, C) is given, then the triangle can be known completely: the other three variables can be found out using the formulae we learnt in this chapter.

There are different cases that arise when a few components of the triangle are given.

Case 1

When three sides (a, b and c) of a triangle are given

The remaining variables can be found by using the following formulae

$\overline{)\begin{array}{cccc}\text{Given}& & & \text{To determine}\\ \\ a,b,c& & & (\mathrm{i})\mathrm{Area}\mathrm{of}\mathrm{\Delta }=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}\\ & & & \text{where},2s=a+b+c\\ \\ \\ & & & (\mathrm{ii})\mathrm{To}\mathrm{find}\mathrm{angles},\mathrm{use}\mathrm{cosine}\mathrm{rule}\\ & & & \cos A=\frac{b^2+c^2-a^2}{2bc},\mathrm{similarly}\mathrm{angle}\mathrm{B}\mathrm{can}\mathrm{be}\mathrm{found}\\ & & & \text{For angle C, use }{180}^0-\text{A}-\text{B}=\text{C}\\ \\ \\ & & & (\mathrm{iii})\mathrm{Angles}\mathrm{can}\mathrm{also}\mathrm{be}\mathrm{calculated}\\ & & & \mathrm{using}\mathrm{half}\mathrm{angle}\mathrm{formula}\\ & & & \tan \frac{A}{2}=\sqrt{\frac{\mathit{(}\mathit{s}\mathit{-}\mathit{b}\mathit{)}\mathit{(}\mathit{s}\mathit{-}\mathit{c}\mathit{)}}{\mathit{s}\mathit{(}\mathit{s}\mathit{-}\mathit{a}\mathit{)}}}\\ \end{array}}$

Problems in which two sides and included angles are given
Case 2

When two sides (say a, b) and included angle (angle C) of a triangle are given.

$\overline{)\begin{array}{cccc}\text{Given}& & & \text{To determine}\\ \\ a,b\text{and}\mathrm{\angle }C& & & (\mathrm{i})\mathrm{the}\mathrm{side}\mathrm{opposite}\mathrm{to}\mathrm{\angle }\mathrm{C}\mathrm{can}\mathrm{be}\mathrm{determined}\\ & & & \text{using cosine rule},i.e.c^2=a^2+b^2-2ab\cos C\\ \\ \\ & & & (\mathrm{ii})\mathrm{using}\mathrm{cosine}\mathrm{rule}\mathrm{again},\mathrm{second}\mathrm{angle}\\ & & & \mathrm{can}\mathrm{also}\mathrm{be}\mathrm{determined},\\ & & & \cos A=\frac{b^2+c^2-a^2}{2bc},\\ \\ \\ & & & (\mathrm{iii})\mathrm{Area}\mathrm{of}\mathrm{\Delta }=\frac{1}{2}\mathrm{a}\cdot \mathrm{b}\sin \mathrm{C}\\ \end{array}}$

Problems in which one side and two angles are given

Case 3

When one side (say a) and two angles (say A and B) are given

$\overline{)\begin{array}{cccc}\text{Given}& & & \text{To determine}\\ \\ a\text{and}\mathrm{\angle }A,\mathrm{\angle }B& & & (\mathrm{i})\mathrm{the}\mathrm{third}\mathrm{angle}\mathrm{is}\mathrm{\angle }\mathrm{C}={180}^0-\mathrm{\angle }\mathrm{A}-\mathrm{\angle }\mathrm{B}\\ \\ \\ & & & (\mathrm{ii})\mathrm{using}\mathrm{sine}\mathrm{rule}\mathrm{side}\mathrm{b}\mathrm{and}\mathrm{c}\mathrm{can}\mathrm{be}\mathrm{found}\\ & & & b=a\frac{\sin B}{\sin A}\text{and}c=a\frac{\sin C}{\sin A}\\ \\ \\ & & & (\mathrm{iii})\mathrm{Area}\mathrm{of}\mathrm{\Delta }=\frac{1}{2}\mathrm{a}\cdot \mathrm{b}\sin \mathrm{C}\\ \end{array}}$

Case 4

When two side a, b and an angle opposite to one of these sides is given (say angle A is given)

Using the sine rule, we get

$\sin \mathrm{B}=\frac{\mathrm{b}}{\mathrm{a}}\sin \mathrm{A}$

Now following possibilities can occur

$\mathbf{1.}\mathbf{When}\frac{\mathbf{b}}{\mathbf{a}}\sin \mathbf{A}\mathbf{>}\mathbf{1}\mathbf{or}\mathbf{a}\mathbf{<}\mathbf{b}\sin \mathbf{A}$

$\mathrm{In}\mathrm{the}\mathrm{relation}\sin \mathrm{B}=\frac{\mathrm{b}}{\mathrm{a}}\sin \mathrm{A},\mathrm{which}\mathrm{means}\sin \mathrm{B}>1,\mathrm{which}\mathrm{is}\mathrm{impossible}$

So, no such triangle exists

$\mathbf{2.}\mathbf{When}\frac{\mathbf{b}}{\mathbf{a}}\sin \mathbf{A}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{then}\sin \mathbf{B}\mathbf{=}\mathbf{1}\Rightarrow \mathrm{\angle }\mathbf{B}\mathbf{=}{\mathbf{90}}^{\mathbf{0}}$

So a unique triangle is possible which is right angle triangle with angle B = 90⁰

$\mathbf{3.}\mathbf{When}\frac{\mathbf{b}}{\mathbf{a}}\sin \mathbf{A}\mathbf{<}\mathbf{1}\mathbf{or}\mathbf{a}\mathbf{>}\mathbf{b}\sin \mathbf{A}$

Here in B < 1, which is possible and hence a triangle will exist

Angle C can be found out using C = 180 - A - B