Question

In right △ABC , the right angle is at C, m∠A=30 degrees , and AC=7√5 units.

What is the perimeter of △ABC ?

Answer 1

`P=(7√(15)+7√(5))\ units`

Explanation:

see the attached figure to better understand the problem

step 1

Find the length side AB (hypotenuse)

we know that

`cos(30^o)=(AC)/(AB)`

substitute

`cos(30^o)=(7√(5))/(AB)`

Remember that

`cos(30^o)=(√(3))/(2)`

so

`(√(3))/(2)=(7√(5))/(AB)`

`AB=(14√(5))/(√(3))`

`AB=(14√(15))/(3)\ units`

step 2

Find the length side BC

`sin(30^o)=(BC)/(AB)`

`sin(30^o)=(1)/(2)`

`(1)/(2)=(BC)/(AB)`

`BC=(AB)/(2)`

substitute the value of AB

`BC=((14√(15))/(3))/(2)`

`BC=(7√(15))/(3)\ units`

step 3

Find the perimeter

`P=AB+BC+AC`

substitute

`P=(14√(15))/(3)+(7√(15))/(3)+7√(5)`

`P=(21√(15))/(3)+7√(5)`

`P=(7√(15)+7√(5))\ units`