Question

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Find the altitude of an isosceles triangle with a vertex angle (that is, the non-base angle) of 70° and a base of 246. Draw a diagram and find the length of the altitude.

175.7
89.5
86.1

Answer 1

@aum

Answer 2

Given: Angle A = 70 degrees. BC = 246
Angle BAD = one half of angle A = 70/2 = 35 degrees.
BD = one half of BC = 246/2 = 123
tan(angle BAD) = opposite / adjacent = BD/AD = 123/AD
AD = 123 / tan(35) = ?

Answer 3

it would equal 86.1

Answer 4

Try again. Make sure the calculator is set to degrees and not radians.

Answer 5

oh ok

Answer 6

I checked but I still got the same answer

Answer 7

@Nnesha

Answer 8

***it would equal 86.1***
you did 123*tan(35)

but that is not what you want to do.
tan(35)= 123/ x (x for unknown)
do you know how to "solve for x" ?

Answer 9

yes, i think. I would have to do .7002 times 123?

Answer 10

@phi

Answer 11

tan(35)= 123/ x

multiply both sides by x:
x * tan(35) = 123/x * x
x * tan(35) = 123
now divide both sides by tan(35)

Answer 12

it gave me 175.7 (rounded)

Answer 13

ok

Answer 14

If you can, can you help me with two more questions?