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two altitudes of an isosceles triangle are equal to 20 cm and 30 cm. determine the possible measures of the base angles of the triangle.

difficulty advanced
 Apr 14, 2015

Best Answer 

 #1
avatar+890 
+11

 

D is the mid-point of AC, AB = BC = a and CE is the perpendicular from C to AB and is of length h.

$$\displaystyle \frac{h}{a}=\sin(\angle EBC) = \sin(180 - 2\angle BAC)=\sin(2\angle BAC)\\ \\ =2\sin(\angle BAC)\cos(\angle BAC)\\ \\=2\times \frac{30}{a}\times \frac{\sqrt{a^{2}-30^{2}}}{a}$$

If h is to equal 20, then, multiplying by a squared and cancelling,

$$\displaystyle a=3\sqrt{a^{2}-900},$$

from which, (square both sides),

$$a=\sqrt{8100/8}\approx 31.8198.$$

 

The base angle can then be calculated. Repeat the calculation for BD = 20, and h = 30.

 Apr 17, 2015
 #1
avatar+890 
+11
Best Answer

 

D is the mid-point of AC, AB = BC = a and CE is the perpendicular from C to AB and is of length h.

$$\displaystyle \frac{h}{a}=\sin(\angle EBC) = \sin(180 - 2\angle BAC)=\sin(2\angle BAC)\\ \\ =2\sin(\angle BAC)\cos(\angle BAC)\\ \\=2\times \frac{30}{a}\times \frac{\sqrt{a^{2}-30^{2}}}{a}$$

If h is to equal 20, then, multiplying by a squared and cancelling,

$$\displaystyle a=3\sqrt{a^{2}-900},$$

from which, (square both sides),

$$a=\sqrt{8100/8}\approx 31.8198.$$

 

The base angle can then be calculated. Repeat the calculation for BD = 20, and h = 30.

Bertie Apr 17, 2015
 #2
avatar+109812 
0

Thanks for doing this one Bertie, it had me totally stumped    

 Apr 17, 2015
 #3
avatar+111401 
0

Nice one, Bertie......!!!!!

 

  

 Apr 17, 2015

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