DAV Class 7 Maths Chapter 8 HOTS
Triangle And Its Properties HOTS
1. In the given figure, AE = AC,\( \angle BAC = 40^\circ , \angle ACF = 75^\circ \) and BCF is a line. Prove that BE = CE.
Answer
\begin{align*} & Given \\ AE &= AC \\ \angle BAC &= 40^\circ \\ \angle ACF &= 75^\circ \\ \\ \text{To prove: } BE & = CE \\ \\ In \ & \triangle AEC \\ AE &= AC \\ \color{magenta} \text{Angles opposite to equal } & \color{magenta} \text{sides of a triangle are equal}\\ \angle AEC &= \angle ACE \\ \\ \angle A + \angle AEC + \angle ACE & = 180^\circ \color{magenta} \text{ (Angle sum property)} \\ 40^\circ + \angle AEC + \angle AEC & = 180^\circ \\ 40^\circ + 2 \angle AEC & = 180^\circ \\ 2 \angle AEC & = 180^\circ - 40^\circ \\ 2 \angle AEC & = 140^\circ \\ \angle AEC & = \frac{140^\circ}{2} \\ \color{green} \angle AEC & = \color{green} 70^\circ \\ \color{green} \angle ACE & = \color{green} 70^\circ \\ \\ \angle A + \angle B & = \angle ACF \color{magenta} \text{ (Exterior angle property)} \\ 40^\circ + \angle B & = 75^\circ \\ \angle B & = 75^\circ - 40^\circ \\ \color{green} \angle B & = \color{green} 35^\circ \\ \\ BCF &\text{ is a straight line} \\ \angle ECB + \angle ACE + \angle ACF & = 180^\circ \\ \angle ECB + 70^\circ + 75^\circ & = 180^\circ \\ \angle ECB + 145^\circ & = 180^\circ \\ \angle ECB & = 180^\circ - 145^\circ \\ \color{green} \angle ECB & = \color{green} 35^\circ \\ \\ Since \implies \angle ECB & = \angle EBC\\ \color{magenta} \text{Sides opposite to equal } & \color{magenta} \text{sides of triangle are equal}\\ \color{green} \therefore BE & = \color{green} CE \\ \end{align*}
2. In the given figure, BO and CO are the bisectors of the exterior angles meeting each other at O. If, \( \angle A = 70^\circ \text{ find } \angle BOC \).
Answer
\begin{align*} In \ \triangle ABC \\ \angle A + \angle ABC + \angle ACB &= 180^\circ \\ 70^\circ + \angle ABC + \angle ACB &= 180^\circ \\ \angle ABC + \angle ACB &= 180^\circ - 70^\circ \\ \color{green} \angle ABC + \angle ACB &= \color{green} 110^\circ \\ \\ \angle CBE & = \angle A + \angle ACB \color{magenta} \text{ (Exterior angle property)} \\ \angle OBE + \angle OBC & = \angle A + \angle ACB \color{magenta} \text{ (OB is the angle bisector of } \angle CBE ) \\ \angle OBC + \angle OBC & = \angle A + \angle ACB \\ \color{green} 2 \angle OBC & = \color{green} \angle A + \angle ACB \\ \\ \angle BCD & = \angle A + \angle ABC \color{magenta} \text{ (Exterior angle property)} \\ \angle OCD + \angle OCB & = \angle A + \angle ABC \color{magenta} \text{ (OC is the angle bisector of } \angle BCD ) \\ \angle OCD + \angle OCB & = \angle A + \angle ABC \\ \color{green} 2 \angle OCB & = \color{green} \angle A + \angle ABC \\ \\ In \ \triangle BOC \\ \angle BOC + \angle OBC + \angle OCB & = 180^\circ \\ \color{green} \angle OBC + \angle OCB & = \color{green} 180^\circ - \angle BOC \\ \\ 2 \angle OBC + 2 \angle OCB & = \angle A + \angle ACB + \angle A + \angle ABC \\ 2 (\angle OBC + \angle OCB) & = \angle A + \angle A + \angle ACB + \angle ABC \\ 2 ( 180^\circ - \angle BOC ) & = 70^\circ + 70^\circ + 110^\circ \\ 2 ( 180^\circ - \angle BOC ) & = 250^\circ \\ 180^\circ - \angle BOC & = \frac{250^\circ}{2} \\ 180^\circ - \angle BOC & = 125^\circ \\ 180^\circ - 125^\circ & = \angle BOC \\ 55^\circ & = \angle BOC \\ \color{green} \angle BOC & = \color{green} 55^\circ \\ \end{align*}
Answer \( \angle BOC = {\color{red} 55^\circ} \)