Mathematics
Jesse27
2015-11-15 03:05:57
What is the approximate length of the base of an isosceles triangle if the congruent sides are 2.5 m and the vertex angle is 25°?
ANSWERS
leahdun
2015-11-15 05:22:11

Let CH be the altitude. CH is also the angle bisector of angle C, so: m(ACH)=m(HCB)=25°/2=12.5°  CH is also a median, so |AH|=|HB| Method 1: by right angle trigonometry, in triangle HBC |HB|=|CB|*sin12.5° (as sine = opposite side / hypotenuse)  |HB|= 2.5 * 0.216 = 0.54 (meters) thus, |AB|=2|HB|=2*0.54 m = 1.08 m Method 2: according to the Cosine law: [latex] |AB|^{2}= |CB|^{2}+ |CA|^{2}-2*|CB|*|CA|*cos(C)[/latex] then substituting the values we know: [latex] |AB|^{2}= (2.5)^{2}+ (2.5)^{2}-2*(2.5)*(2.5)*cos25[/latex] [latex] |AB|^{2}= 2*(2.5)^{2}-2*(2.5)^{2}(0.906)[/latex] [latex] |AB|^{2}= 2*(2.5)^{2}(1-0.906)[/latex] [latex] |AB|^{2}= (2.5)^{2}(0.188)[/latex] taking the square root of both sides: [latex]|AB|= 2.5* sqrt{0.2}=2.5*0.43358=1.08[/latex]   (meters) Answer: 1.08 m

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