Question

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

Answer 1

Answer:

$P=(7\sqrt{15}+7\sqrt{5})\ units$

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the length side AB (hypotenuse)

we know that

$cos(30^o)=\frac{AC}{AB}$

substitute

$cos(30^o)=\frac{7\sqrt{5}}{AB}$

Remember that

$cos(30^o)=\frac{\sqrt{3}}{2}$

so

$\frac{\sqrt{3}}{2}=\frac{7\sqrt{5}}{AB}$

$AB=\frac{14\sqrt{5}}{\sqrt{3}}$

$AB=\frac{14\sqrt{15}}{3}\ units$

step 2

Find the length side BC

$sin(30^o)=\frac{BC}{AB}$

$sin(30^o)=\frac{1}{2}$

$\frac{1}{2}=\frac{BC}{AB}$

$BC=\frac{AB}{2}$

substitute the value of AB

$BC=\frac{\frac{14\sqrt{15}}{3}}{2}$

$BC=\frac{7\sqrt{15}}{3}\ units$

step 3

Find the perimeter

$P=AB+BC+AC$

substitute

$P=\frac{14\sqrt{15}}{3}+\frac{7\sqrt{15}}{3}+7\sqrt{5}$

$P=\frac{21\sqrt{15}}{3}+7\sqrt{5}$

$P=(7\sqrt{15}+7\sqrt{5})\ units$