document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "Please help, i will need them for Wednesday 27th March. thankyou \n" ); document.write( "

Algebra.Com's Answer #446041 by KMST(5328) $\"\"$ $\"About$

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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.
\n" ); document.write( "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
\n" ); document.write( " $\"xh=15cm%5E2\"$
\n" ); document.write( "The measures of angles ABP and CBP are the same, and half of the measure of ABC, meaning
\n" ); document.write( " $\"24%5Eo%2F2=12%5Eo\"$
\n" ); document.write( "The trigonometric ratios, applied to ABP tell us that
\n" ); document.write( " $\"x%2FAB=sin%2812%5Eo%29\"$ --> $\"x=AB%2Asin%2812%5Eo%29\"$
\n" ); document.write( " $\"h%2FAB=cos%2812%5Eo%29\"$ --> $\"h=AB%2Acos%2812%5Eo%29\"$ \r
\n" ); document.write( "\n" ); document.write( "So $\"xh=AB%5E2%2Asin%2812%5Eo%29%2Acos%2812%5Eo%29\"$ --> $\"AB%5E2%2Asin%2812%5Eo%29%2Acos%2812%5Eo%29=15\"$ --> $\"AB%5E2=15%2Fsin%2812%5Eo%29%2Acos%2812%5Eo%29\"$ --> $\"AB%5E2=73.7578\"$ --> $\"AB=sqrt%2873.7578%29\"$ --> $\"AB=8.588\"$ cm rounded to $\"8.6cm\"$
\n" ); document.write( " $\"x=AB%2Asin%2812%5Eo%29\"$ --> $\"x=8.588%2Asin%2812%5Eo%29\"$ --> $\"x=1.7856cm\"$ (rounded
\n" ); document.write( "and $\"AC=2x=1.7856%2A2=3.57cm\"$ rounded to $\"3.6cm\"$
\n" ); document.write( "
\n" ); document.write( "Alternatively, we could have used the fact that the area of a triangle can be calculated as
\n" ); document.write( " $\"%28AB%29%28BC%29sin%28ABC%29%2F2\"$ and since $\"AB=BC\"$ , then the area is $\"AB%5E2%2Asin%2824%5Eo%29%2F2\"$
\n" ); document.write( " $\"AB%5E2%2Asin%2824%5Eo%29%2F2=15\"$ --> $\"ab%5E2=2%2A15%2Fsin%2824%5Eo%29\"$ --> $\"AB%5E2=73.7578\"$
\n" ); document.write( "
\n" ); document.write( "Q2. Using the same drawing, the length of both legs of the isosceles triangle (AB and BC) is the same, $\"5.7cm\"$ .
\n" ); document.write( "As above the area of the triangle can be calculated as
\n" ); document.write( " $\"%28AB%29%28BC%29sin%28ABC%29%2F2\"$ and since $\"AB=BC\"$ , then the area is $\"AB%5E2%2Asin%28ABC%29%2F2\"$ , so
\n" ); document.write( " $\"5.7%5E2%2Asin+ABC%2F2=4\"$ --> $\"sin+ABC=2%2A4%2F5.7%5E2\"$ --> $\"sin+ABC=0.24623\"$ (rounded)
\n" ); document.write( "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 $\"14.2545%5Eo%2F2=7.1273%5Eo\"$
\n" ); document.write( "and the measure of CBA $\"90%5Eo-7.1273%5Eo=82.8727%5Eo\"$ .
\n" ); document.write( "Then the base angles would be $\"82.9%5Eo\"$ and the vertex angle would be $\"14.3%5Eo\"$
\n" ); document.write( "
\n" ); document.write( "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 measuring $\"2%2A82.8727%5Eo=165.7%5Eo\"$ } which also has $\"sin%2814.2545%5Eo%29=0.24623\"$ \n" ); document.write( "

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