Question

in a triangle abc angle a equal to 30 degree angle B equal to 60 degree ab is equal to ten centi metres then the length of the shortest trisector of angle C​
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Answer 1

Step-by-step explanation:

Give in ∆ABC, A=30⁰, B=68⁰

Then c=90⁰

Now From Sine rule of triangle we get,

a,=2R sin A ,b=2R sin B and C=2Rsin

[where R is the radius of the circum circle of ∆ABC,]

= sin A : Sin B : sin C :

= sin 30⁰: Sin 60⁰: Sin 90⁰:

Answer 2

Length of the shortest trisector of angle C​ is  5√3 /2

Given:

• ΔABC
• AB = 10 cm
• ∠A = 30°
• ∠B = 60°

To Find:

• Length of the shortest trisector of angle C​

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Solution:

• Sinθ = Opposite side/Hypotenuse  
• Cosθ = Adjacent side/ Hypotenuse
• Sum of angle of a triangle is 180°

Refer the figure for Better Understanding

Step 1:

Using sum of angles of triangle

∠A + ∠B + ∠C = 180°

=> 30° + 60° + ∠C = 180°

=> ∠C = 90°

Hence Δ ABC is Right angle

Step 2:

Using sin and cos

sin30° = BC/AB = BC/10

=> 1/2 = BC/10

=> BC = 5 cm

cos30° = AC/AB = BC/10

=> 1/2 = BC/10

=> BC = 5 cm

Step 3:

CE and CF are trisector of ∠C hence

∠BCE = ∠ECF = ∠ACF = 30°

∠BCF = ∠BCE +  ∠ECF = 60°

Now ΔBEC and ΔFEC are right angle triangle

and ΔBCF is equilateral triangle ( as ∠BCF = 60° and ∠B = 60°)

Step 4:

As ΔBCF is equilateral triangle Hence

CF = BC = 5 cm

CE = 5√3 /2    as CE is altitude of ΔBCF

5√3 /2 < 5

Hence length of the shortest trisector of angle C​ is  5√3 /2