Hello everyone,
I’ve come across a question where I need to find the perimeter of a right angle triangle given its area and three sides (the only angle written in the picture is 40 degrees, but the other must be 50 degrees given that it is a right triangle). I honestly am at a complete loss at how to solve this, I’d appreciate any help.
Thanks!
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$\begingroup$Since its a right-angled triangle, follow these 4 steps to get the solution:
Step-1: Angle is given use the Trigonometric relation of Tangent:
tan(40) = Height/Base ------eq(1)Step-2: Triangle area formula i.e.
1/2 * Height * Base = 20 ------eq(2)Now you can substitute one unknown variable from eq(1) to the eq(2) or vice-versa to obtain both the Base and Height of the triangle.
Step-3: Use the Pythagoras theorem
Hypotenus^2 = Height^2 + Base^2Step-4: Perimeter = Hypotenus + Height + Base
$\endgroup$ 2 $\begingroup$In $\triangle ABC,$ with $\angle ACB=90^o$ and $\angle BAC=40^o$: Let $a=BC,\,b=CA,\,c=AB.$ We have $$20 \text { cm}^2=ab/2=(c \cos 40^o)(c\sin 40^o)/2.$$ Therefore $c=\sqrt {\frac {40}{ (\cos 40^o)(\sin 40^o)}}\,$cm.
Knowing $c,$ we have $a=c\cos 40^o$ and $b=c\sin 40^0$.
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