Question

Hey everyone !

Answer 1

Answer:

The measure of the largest angle is 120°

Step-by-step explanation:

Lets explain how to find the measure of an angle from the length of the

sides of the triangle

- We can do that by using the cosine rule

- If the three angles of the triangle are A , B , C, then the side opposite

 to angle A is BC , the side opposite to angle B is AC and the side

 opposite to angle C is AB, So to find measure of angle A use the rule

 $cos(A)=\frac{(AB)^{2}+(AC)^{2}-(BC)^{2}}{2(AB)(AC)}$

Lets solve the problem

- Assume that the triangle is ABC where AB = 14 cm , BC = 10 cm and

 AC = 6 cm

- We need to find the measure of the largest angle

- The largest angle is opposite to the longest side

∵ The longest side is AB

∴ The largest angle is C

By using the rule above

∴ $cos(C)=\frac{(AC)^{2}+(BC)^{2}-(AB)^{2}}{2(AC)(BC)}$

∵ AB = 14 cm , BC = 10 cm , AC = 6 cm

∴ $cos(C)=\frac{(6)^{2}+(10)^{2}-(14)^{2}}{2(6)(10)}$

∴ $cos(C)=\frac{36+100-196}{120}$

∴ $cos(C)=\frac{-60}{120}=-0.5$

∴ cos(C) = -0.5 ⇒ that means angle C is obtuse angle

∴ m∠C = $cos^{-1}(-0.5)=120$

* The measure of the largest angle is 120°

Answer 2

Hello!

A triangle has sides measuring 14 cm, 10 cm and 6 cm. The measure of the largest angle of this triangle measures:

Note: Perimeter = 30

We have the following data:  

p (perimeter) = 14 + 10 + 6 = 30

a = 14 cm  

b = 10 cm  

c = 6 cm  

α (angle) = ?  

*** Note: The largest angle (α) is always opposite the larger side.

We apply the data to the Cosine Law, let's see:

$a^2 = b^2 + c^2 - 2*b*c*cos\:\alpha$

$14^2 = 10^2 + 6^2 - 2*10*6*cos\:\alpha$

$196 = 100 + 36 - 120\:cos\:\alpha$

$120\:cos\:\alpha = -196 + 100 + 36$

$120\:cos\:\alpha = -60$

$cos\:\alpha = \dfrac{-6\diagup\!\!\!\!0}{12\diagup\!\!\!\!0} \frac{\div6}{\div6}$

$cos\:\alpha = \dfrac{-1}{2}$

$\boxed{\boxed{cos\:\alpha = 120\º}}\:\:\:\:\:\:\bf\purple{\checkmark}$

Answer:

The measure of the largest angle of the triangle is 120º

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