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
Answers
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:
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
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
