Question

In the figure B=90', A = C and BC=10cm, then find the value of tan 45º.
A
B
10cm​

Answer 1

Step-by-step explanation:

In right ΔBCD:

cot45°=BDCD=1(1)

In right ΔACD:

cot60°=ADCD=13–√(2)

Adding equations (1) and (2) :

BD+ADCD=1+13–√

ABCD=3–√+13–√

10CD=3–√+13–√

CD=103–√3–√+1

CD=103–√(3–√−1)2

CD=5(3−3–√)

Answer 2

Answer:

tan 45° =1

Step-by-step explanation:

Let ∠A=∠C=x

Then x+x+90=180

2x=180°-90°

2x=90°

x=90°/2=45°

tan C= AB/AC

tan 45° =10/10=1