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Question: How do you find the largest and smallest angle in a triangle with sides \(3,6\) and \(7?\)...

How do you find the largest and smallest angle in a triangle with sides 3,63,6 and 7?7?

Explanation

Solution

In this question, we are going to find the largest and smallest angle in a triangle for the given sides.
In this we have the values of the three sides of the triangle namely 3,63,6 and 77
First, we have to use the law of cosine to find one of the angles (angle aa).
Next we have to find another side (angle cc) by again using the law of cosine.
Finally, we can find angle bb by using angles of a triangle add to 180∘{180^ \circ }
By solving the triangle we can get the required result.

Formula used: The law of cosine is defined by
a2=b2+c2−2bc×cos⁡A{a^2} = {b^2} + {c^2} - 2bc \times \cos A

Complete step-by-step solution:
In this question, we are going to find the largest and the smallest angle of a triangle.
In this given the values of the three sides of a triangle namely 3,63,6 and 77
Here a=3, b=6, c=7a = 3,\,b = 6,\,c = 7
First we are going to find the smallest angle (angle aa).
Applying law of cosine we get,
⇒(3)2=(6)2+(7)2−2(6)(7)×cos⁡A\Rightarrow {\left( 3 \right)^2} = {\left( 6 \right)^2} + {\left( 7 \right)^2} - 2\left( 6 \right)\left( 7 \right) \times \cos A
On simplify the term and we get,
⇒9=36+49−2×42×cos⁡A\Rightarrow 9 = 36 + 49 - 2 \times 42 \times \cos A
On adding and multiply the term and we get
⇒9=85−84cos⁡A\Rightarrow 9 = 85 - 84\cos A
On rewriting we get,
⇒84cos⁡A=85−9\Rightarrow 84\cos A = 85 - 9
Let us subtract the term and we get
⇒84cos⁡A=76\Rightarrow 84\cos A = 76
Let us divide the term and we get
⇒cos⁡A=7684\Rightarrow \cos A = \dfrac{{76}}{{84}}
On dividing the term and we get
⇒cos⁡A=1921\Rightarrow \cos A = \dfrac{{19}}{{21}}
Then we get,
⇒cos⁡A=0.90476\Rightarrow \cos A = 0.90476
On rewriting we get,
⇒A=cos⁡−1(0.90476)\Rightarrow A = {\cos ^{ - 1}}(0.90476)
Then we get,
A=25.84∘A = 25.84^\circ
Next we have to find the largest angle cc by again using the law of cosine.
⇒c2=a2+b2−2ab×cos⁡C\Rightarrow {c^2} = {a^2} + {b^2} - 2ab \times \cos C
On putting the values and we get
⇒(7)2=(3)2+(6)2−2(3)(6)×cos⁡C\Rightarrow {\left( 7 \right)^2} = {\left( 3 \right)^2} + {\left( 6 \right)^2} - 2\left( 3 \right)\left( 6 \right) \times \cos C
On squaring the term and we get
⇒49=9+36−36×cos⁡C\Rightarrow 49 = 9 + 36 - 36 \times \cos C
On adding the term and we get,
⇒49=45−36cos⁡C\Rightarrow 49 = 45 - 36\cos C
On rewriting we get,
⇒36cos⁡C=45−49\Rightarrow 36\cos C = 45 - 49
Let us subtract the term and we get
⇒36cos⁡C=−4\Rightarrow 36\cos C = - 4
On dividing the term and we get
⇒cos⁡C=−436\Rightarrow \cos C = \dfrac{{ - 4}}{{36}}
On cancel, we get
⇒cos⁡C=−19\Rightarrow \cos C = \dfrac{{ - 1}}{9}
Then we get,
⇒cos⁡C=−0.11\Rightarrow \cos C = - 0.11
⇒C=cos⁡−1(−0.11)\Rightarrow C = {\cos ^{ - 1}}( - 0.11)
⇒C=96.38\Rightarrow C = 96.38Degree
Now we are going to find the angle bb,
Angle B=180−A−CB = 180 - A - C
Let us putting we get
⇒B=180−12.53−96.38\Rightarrow B = 180 - 12.53 - 96.38
On adding the term and we get
⇒B=180−108.91\Rightarrow B = 180 - 108.91
Let us subtract the term and we get
⇒B=71.09\Rightarrow B = 71.09
The angle of BB is 71.09∘71.09^\circ
Now we have completely solved the triangle, that is we have found all its angles.

Here the largest angle is 96.3896.38 degree and the smallest angle is 25.8425.84 degree

Note: The angle opposite the smallest side of a triangle has the smallest measure. Likewise, the angle opposite the largest side has the largest measure. So, if given three side lengths, in order to put the angles in order from smallest to largest.