Question

Please answer my question please ☹️

Answer 1

14) Draw a triangle ABC, and then draw a straight line DE, touching angle A (top portion), now draw another straight line QP, touching the base BC.

Now, extend the sides BA and CA into BZ and CY respectively,then draw exterior angles named angle ABQ and angle ACP.

Now, we have two parallel lines and a transversal.

Now, angle ABQ=angle ZAE (alternate angles)

angle BAC=angle YAZ (vertically opposite angles)

angle ACP=YAD (alternate angles)

Now angle YAZ, angle YAD and angle ZAE make a straight line, and we know that a straight line=180°

Hence, proved.

15) Draw a triangle ABC and let angle A=110°, so let the other two angles be x°.

So, by angle sum property, we can solve this easily.

angle A+angle B+angle C=180°

110°+x°+x°=180°

2x°=180°-110°

2x°=70°

x°=70°/2°

x°=35°

So, the other two angles are 35°

So, 35°+110°+35°=180°

So, the answer is correct.

16) 1) False because Triangle cannot have more than one obtuse angle because there are 3 side and if you add three sides it equals to 180 degree andobtuse angle is greater than 90 degree and the other two has to be less than 90 degree. 

2) False as Not only can you have such a triangle, but you can't have a triangle with fewer than two acute angles.

3) False as If all the angles are less than 60° , then theirsum cannot be 180°. Therefore , A triangle with all the 3 angles less than 60° is invalid.