Question Video: Identifying the Type of an Angle in a Triangle by Applying Triangular Inequalities | Nagwa Question Video: Identifying the Type of an Angle in a Triangle by Applying Triangular Inequalities | Nagwa

# Question Video: Identifying the Type of an Angle in a Triangle by Applying Triangular Inequalities Mathematics • Second Year of Preparatory School

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In triangle πππ, (ππ)Β² > (ππ)Β² β (ππ)Β². What type of angle is π?

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### Video Transcript

In triangle πππ, ππ squared is greater than ππ squared minus ππ squared. What type of angle is π?

If we consider any triangle πππ, there are three possibilities for angle π. The first diagram shows that angle π is a right angle. The second diagram shows angle π is an acute angle, as it is less than 90 degrees or a right angle. In the third diagram, angle π is an obtuse angle, as it is greater than 90 degrees but less than 180 degrees. We need to consider the relationship between the lengths in the triangle, ππ, ππ, and ππ. In this question, we need to decide which triangle corresponds to the inequality ππ squared is greater than ππ squared minus ππ squared.

Pythagorasβ theorem states that in any right-angled triangle, π squared plus π squared is equal to π squared, where π is the length of the longest side known as the hypotenuse. This means that in our first diagram, ππ squared plus ππ squared is equal to ππ squared. As angle π gets smaller, the length ππ also gets smaller. This means that in the second diagram, the sum of the other two sides squared will be greater than ππ squared. Conversely, as angle π gets larger, the length ππ gets larger. This means that in our third diagram, ππ squared plus ππ squared is less than ππ squared.

We can now work out which of our three triangles matches the initial inequality. We can rule out the answer right angle as this has an equation and not an inequality. Letβs consider the initial inequality ππ squared is greater than ππ squared minus ππ squared. Adding ππ squared to both sides of this inequality gives us ππ squared plus ππ squared is greater than ππ squared. This corresponds to the second diagram. We can therefore conclude that, in this case, angle π is acute. It is less than 90 degrees.

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