Question

In a right triangle abc , right angled at C , if angle B =60 degree and ab =15 units find remaining angles and sides

Answer 1

Answer:

∠A=30° and AC and BC is 13 and 7.5 units respectively

Step-by-step explanation:

Given that in a right triangle ABC , right angled at C , if angle B =60 degree and AB =15 units

we have to find the remaining angles.

By angle sum property,  sum of angles in right angle is 180°

∠A+∠B+∠C=180°

∠A+ 60°+90°=180

∠A=180°-150°=30°

∠A=30°

As AB= 15

Now by trigonometric ratios we find the other sides of triangle ABC

$\sin A=\frac{Perpendicular}{Hypotenuse}=\frac{CB}{AB}$

$\sin 30=\frac{CB}{15}$

$CB=15\times \frac{1}{2}=7.5 units$

$\sin B=\frac{Perpendicular}{Hypotenuse}=\frac{AC}{AB}$

$\sin 60=\frac{AC}{15}$

$AC=15\times \frac{\sqrt3}{2}=12.99 units\sim 13 units$

Hence, ∠A=30° and AC and BC is 13 and 7.5 units

   

Answer 2

solution:

Given : In a right ∆ ABC,

∠C = 90°,

if ∠B = 60° and

AB = 15 units.

In ∆ABC,  

∠A + ∠B + ∠C = 180°

[Sum of angles in a ∆ = 180°]

∠A + 60° + 90° = 180°  

∠A + 150°  = 180°  

∠A = 180° - 150°

∠A = 30°  

In ∆ABC,  

With reference to ∠B ,  

Base = BC , Perpendicular = AC , Hypotenuse = AB

cos 60° = B/ H = BC/AB

½ = BC/15

2BC = 15  

BC = 15/2

BC = 7.5 units

sin 60° = P/H = AC/AB  

√3/2 = AC/15

2AC = 15√3

AC = (15√3) /2 units

Hence, ∠A = 30°, BC = 7.5 units ,AC = (15√3) /2 units

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Hope this helps!