Asked by Nkasiobi
In isosceles triangle ACB, AB=AC =4cm,BAC=30 and CN is an altitude. Find BN
Please I need the diagram
Please I need the diagram
Answers
Answered by
Tammy
Draw 2 arms AB and AC which form an angle of 30°. Join BC
draw a line from A to BC, to meet BC at N at 90°
look at triangle ABN,
cos 15° = BN/4
BN = 4sin15° = appr 1.035
so BC = 2(1.035..) = appr 2.07
or
you could use the cosine law
BC^2 = 4^2 + 4^2 - 2(4)(4)cos30 = 4.28718..
BC = √4.28718.. = appr 2.07 , same as before
draw a line from A to BC, to meet BC at N at 90°
look at triangle ABN,
cos 15° = BN/4
BN = 4sin15° = appr 1.035
so BC = 2(1.035..) = appr 2.07
or
you could use the cosine law
BC^2 = 4^2 + 4^2 - 2(4)(4)cos30 = 4.28718..
BC = √4.28718.. = appr 2.07 , same as before
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