The length of $A D$ (in cm.) is

10

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100

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75

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50

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Solution

The correct option is D $50$
In $△ A B C$
$B C = A C$
Thus, $∠ A B C = ∠ A C B = x$ (Isosceles triangle Property)
Now, $∠ A C D = ∠ A B C + ∠ A C B$ (Exterior angle property)
$60 = x + x$
$x = 30^{\circ}$
Now, $sin ∠ A B C = \frac{P}{H}$
$sin 30 = \frac{A D}{A B}$
$\frac{1}{2} = \frac{A D}{100}$
$A D = 50$ cm

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The length of AD (in cm.) is

The correct option is D 50In △ABCBC=ACThus, ∠ABC=∠ACB=x (Isosceles triangle Property)Now, ∠ACD=∠ABC+∠ACB (Exterior angle property)60=x+xx=30∘Now, sin∠ABC=PHsin30=ADAB12=AD100AD=50 cm

The length of $A D$ (in cm.) is