Question

Is it possible to have a triangle, in which
(i) two of the angles are right?
(ii) two of the angles are obtuse?
(iii) two of the angles are acute?
(iv) each angle is less than 60°?
(v) each angle is greater than 60°?
(vi) each angle is equal to 60°?
Give reasons in support of your answer in each case.

Answer 1

(i) No, because if there are two right angles in a triangle, then the third angle of the triangle must be zero, which is not possible.
(ii) No, because as we know that the sum of all three angles of a triangle is always 180$°$. If there are two obtuse angles, then their sum will be more than 180$°$, which is not possible in case of a triangle.
(iii) Yes, in right triangles and acute triangles, it is possible to have two acute angles.
(iv) No, because if each angle is less than 60$°$, then the sum of all three angles will be less than 180$°$, which is not possible in case of a triangle.
Proof :
$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{be}\angle \mathrm{A},\angle \mathrm{B}\mathrm{and}\angle \mathrm{C}. \\ \mathrm{As}\mathrm{per}\mathrm{the}\mathrm{given}\mathrm{information}, \\ \angle \mathrm{A}<60°...\left(\mathrm{i}\right) \\ \angle \mathrm{B}<60°...\left(\mathrm{ii}\right) \\ \angle \mathrm{C}<60°...\left(\mathrm{iii}\right) \\ \mathrm{On}\mathrm{adding}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{get}: \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}<60°+60°+60° \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}<180° \\ \mathrm{We}\mathrm{can}\mathrm{see}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{three}\mathrm{angles}\mathrm{is}\mathrm{less}\mathrm{than}180°,\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{possible}\mathrm{for}\mathrm{a}\mathrm{triangle}. \\ \mathrm{Hence},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}\mathrm{it}\mathrm{is}\mathrm{not}\mathrm{possible}\mathrm{for}\mathrm{each}\mathrm{angle}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{to}\mathrm{be}\mathrm{less}\mathrm{than}60°. \end{gathered}$$

(v) No, because if each angle is greater than 60$°$, then the sum of all three angles will be greater than 180$°$, which is not possible.
Proof :
$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{triangle}\mathrm{be}\angle \mathrm{A},\angle \mathrm{B}\mathrm{and}\angle \mathrm{C}. \\ \mathrm{As}\mathrm{per}\mathrm{the}\mathrm{given}\mathrm{information}, \\ \angle \mathrm{A}>60°...\left(\mathrm{i}\right) \\ \angle \mathrm{B}>60°...\left(\mathrm{ii}\right) \\ \angle \mathrm{C}>60°...\left(\mathrm{iii}\right) \\ \mathrm{On}\mathrm{adding}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{get}: \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}>60°+60°+60° \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}>180° \\ \mathrm{We}\mathrm{can}\mathrm{see}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{triangle}\mathrm{are}\mathrm{greater}\mathrm{than}180°,\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{possible}\mathrm{for}\mathrm{a}\mathrm{triangle}. \\ \mathrm{Hence},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}\mathrm{it}\mathrm{is}\mathrm{not}\mathrm{possible}\mathrm{for}\mathrm{each}\mathrm{angle}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{to}\mathrm{be}\mathrm{greater}\mathrm{than}60°. \end{gathered}$$

(vi) Yes, if each angle of the triangle is equal to 60$°$, then the sum of all three angles will be 180$°$, which is possible in case of a triangle.
Proof :
$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{triangle}\mathrm{be}\angle \mathrm{A},\angle \mathrm{B}\mathrm{and}\angle \mathrm{C}. \\ \mathrm{As}\mathrm{per}\mathrm{the}\mathrm{given}\mathrm{information}, \\ \angle \mathrm{A}=60°...\left(\mathrm{i}\right) \\ \angle \mathrm{B}=60°...\left(\mathrm{ii}\right) \\ \angle \mathrm{C}=60°...\left(\mathrm{iii}\right) \\ \mathrm{On}\mathrm{adding}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{get}: \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=60°+60°+60° \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180° \\ \mathrm{We}\mathrm{can}\mathrm{see}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{triangle}\mathrm{is}\mathrm{equal}\mathrm{to}180°,\mathrm{which}\mathrm{is}\mathrm{possible}\mathrm{in}\mathrm{case}\mathrm{of}\mathrm{a}\mathrm{triangle}. \\ \mathrm{Hence},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}\mathrm{it}\mathrm{is}\mathrm{possible}\mathrm{for}\mathrm{each}\mathrm{angle}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{to}\mathrm{be}\mathrm{equal}\mathrm{to}60°. \end{gathered}$$