SOLUTION: Q1. The area of this isosceles triangle is 15cm squared. The angle ABC is 24 degrees. Work out both lengths correct to 1 decimal place. Q2. This isosceles triangle as sides of l

Algebra ->  Trigonometry-basics -> SOLUTION: Q1. The area of this isosceles triangle is 15cm squared. The angle ABC is 24 degrees. Work out both lengths correct to 1 decimal place. Q2. This isosceles triangle as sides of l      Log On


   

Question 729425: Q1. The area of this isosceles triangle is 15cm squared. The angle ABC is 24 degrees. Work out both lengths correct to 1 decimal place.
Q2. This isosceles triangle as sides of length 5.7cm and area 4cm squared. The diagram shows that there are 2 possible answers for the angle x. Work out both angles correct to 1 decimal place.
Please help, i will need them for Wednesday 27th March. thankyou
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Q1. 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 xcm . The length of altitude BP is the height hcm and the area of the triangle is
xh=15cm%5E2
The measures of angles ABP and CBP are the same, and half of the measure of ABC, meaning
24%5Eo%2F2=12%5Eo
The trigonometric ratios, applied to ABP tell us that
x%2FAB=sin%2812%5Eo%29 --> x=AB%2Asin%2812%5Eo%29
h%2FAB=cos%2812%5Eo%29 --> h=AB%2Acos%2812%5Eo%29
So xh=AB%5E2%2Asin%2812%5Eo%29%2Acos%2812%5Eo%29 --> AB%5E2%2Asin%2812%5Eo%29%2Acos%2812%5Eo%29=15 --> AB%5E2=15%2Fsin%2812%5Eo%29%2Acos%2812%5Eo%29 --> AB%5E2=73.7578 --> AB=sqrt%2873.7578%29 --> AB=8.588cm rounded to 8.6cm
x=AB%2Asin%2812%5Eo%29 --> x=8.588%2Asin%2812%5Eo%29 --> x=1.7856cm (rounded
and AC=2x=1.7856%2A2=3.57cm rounded to 3.6cm

Alternatively, we could have used the fact that the area of a triangle can be calculated as
%28AB%29%28BC%29sin%28ABC%29%2F2 and since AB=BC , then the area is AB%5E2%2Asin%2824%5Eo%29%2F2
AB%5E2%2Asin%2824%5Eo%29%2F2=15 --> ab%5E2=2%2A15%2Fsin%2824%5Eo%29 --> AB%5E2=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
%28AB%29%28BC%29sin%28ABC%29%2F2 and since AB=BC , then the area is AB%5E2%2Asin%28ABC%29%2F2, so
5.7%5E2%2Asin+ABC%2F2=4 --> sin+ABC=2%2A4%2F5.7%5E2 --> sin+ABC=0.24623 (rounded)
Since sin%2814.2545%5Eo%29=0.24623 it could be that angle ABC measures 14.2545%5Eo, which would make the measure of APB 14.2545%5Eo%2F2=7.1273%5Eo
and the measure of CBA 90%5Eo-7.1273%5Eo=82.8727%5Eo .
Then the base angles would be 82.9%5Eo and the vertex angle would be 14.3%5Eo

However, the right triangles could be reversed, with a base angle CBA measuring 7.1%5Eo} and ABP measuring 82.8727%5Eo for a vertex angle measuring2%2A82.8727%5Eo=165.7%5Eo} which also has sin%2814.2545%5Eo%29=0.24623