Question 729425
Q1. {{{drawing(300,300,-5,5,-1,9,
green(triangle(0,0,0,8,4,0)),locate(-0.1,-0.1,P),
triangle(0,0,0,8,-4,0),rectangle(-0.3,0,0,0.3),
locate(-4.3,0.2,A),locate(-0.1,8.5,B),
locate(4.1,0.2,C), locate(0.1,4.5,h),
locate(-2.1,0.5,x),locate(1.9,0.5,x)
)}}} If B is the vertex, the point where the two equal length sides meet, the altitude to the vertex,splits the triangle into two congruent right triangles.
Half the length of base AC is {{{x}}}{{{cm}}} . The length of altitude BP is the height {{{h}}}{{{cm}}} and the area of the triangle is
{{{xh=15cm^2}}}
The measures of angles ABP and CBP are the same, and half of the measure of ABC, meaning
{{{24^o/2=12^o}}}
The trigonometric ratios, applied to ABP tell us that
{{{x/AB=sin(12^o)}}} --> {{{x=AB*sin(12^o)}}}
{{{h/AB=cos(12^o)}}} --> {{{h=AB*cos(12^o)}}}

So {{{xh=AB^2*sin(12^o)*cos(12^o)}}} --> {{{AB^2*sin(12^o)*cos(12^o)=15}}} --> {{{AB^2=15/sin(12^o)*cos(12^o)}}} --> {{{AB^2=73.7578}}} --> {{{AB=sqrt(73.7578)}}} --> {{{AB=8.588}}}cm rounded to {{{8.6cm}}}
{{{x=AB*sin(12^o)}}} --> {{{x=8.588*sin(12^o)}}} --> {{{x=1.7856cm}}} (rounded
and {{{AC=2x=1.7856*2=3.57cm}}} rounded to {{{3.6cm}}}
 
Alternatively, we could have used the fact that the area of a triangle can be calculated as
{{{(AB)(BC)sin(ABC)/2}}} and since {{{AB=BC}}} , then the area is {{{AB^2*sin(24^o)/2}}}
{{{AB^2*sin(24^o)/2=15}}} --> {{{ab^2=2*15/sin(24^o)}}} --> {{{AB^2=73.7578}}}
 
Q2. Using the same drawing, the length of both legs of the isosceles triangle (AB and BC) is the same, {{{5.7cm}}}.
As above the area of the triangle can be calculated as
{{{(AB)(BC)sin(ABC)/2}}} and since {{{AB=BC}}} , then the area is {{{AB^2*sin(ABC)/2}}}, so
{{{5.7^2*sin ABC/2=4}}} --> {{{sin ABC=2*4/5.7^2}}} --> {{{sin ABC=0.24623}}} (rounded) 
Since {{{sin(14.2545^o)=0.24623}}} it could be that angle ABC measures {{{14.2545^o}}}, which would make the measure of APB {{{14.2545^o/2=7.1273^o}}}
and the measure of CBA {{{90^o-7.1273^o=82.8727^o}}} .
Then the base angles would be {{{82.9^o}}} and the vertex angle would be {{{14.3^o}}}
 
However, the right triangles could be reversed, with a base angle CBA measuring {{{7.1^o}}}} and ABP measuring {{{82.8727^o}}} for a vertex angle measuring{{{2*82.8727^o=165.7^o}}}} which also has {{{sin(14.2545^o)=0.24623}}}