Question

In ΔABC, ∠B = 90, find the measure of the parts of the triangle other than the ones which are given: M∠A=30, AC=10.

Answer 1

SOLUTION :

Given :

In ∆ ABC ,∠B = 90°, ∠A = 30°, AC= 10

∠A + ∠B + ∠C = 180°

30° + 90° + ∠C = 180°

120° + ∠C = 180°

∠C = 180° - 120°

∠C = 60°

For sinC , base = BC , Perpendicular = AB, hypotenuse = AC

sinC = Perpendicular/ hypotenuse = AB/AC

sin 60 = AB/AC

√3/2 = AB/10

2AB = 10√3

AB = 10√3/2 = 5√3

AB = 5√3

For sinA , base = AB , Perpendicular = BC, hypotenuse = AC

sinA = Perpendicular/ hypotenuse = BC/AC

sin 30 = BC/AC

1/2 = BC/10

2BC = 10

BC = 10/2 = 5

AB = 5

Hence, ∠C = 60° , AB = 5√3 , AB = 5

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

It is given that ,

In ∆ABC , <B = 90°

<A = 30° ,

AC = 10

i ) Sin30° = BC/AC

1/2 = BC/10

( 10/2 ) = BC

BC = 5

ii ) Cos 30° = AB/AC

√3/2 = AB/10

( √3 × 10 )/2 = AB

5√3 = AB

AB = 5√3

Therefore ,

Measurements of other parts of ∆ are

AB = 5√3 ,

BC = 5

I hope this helps you.

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