AB=10cm BC= 5cm
AC=5 root 3 cm
Find tan a . tan b
Answers
Answer:
as pythagoras theorm is applicable so it is right angle triangle right angled at c
tan.a = 5root 3/5
tan.b = 5/5root3
tan.a . tanb = 1
Answer:
The value of \bold{\sin A=\frac{5}{13}, \ {Tan} A=\frac{5}{12}, \sin C=\frac{12}{13}, \quad \cot C=\frac{5}{13}}sinA=
13
5
, TanA=
12
5
,sinC=
13
12
,cotC=
13
5
.
Solution:
The triangle ABC is drawn below which is
In the triangle ABC angle B is 90 degree. The “length of the side AB” = 12cm and the “length of the side BC” = 5cm.
Now to find the “length of side AC” we use Pythagoras theorem we get A B^{2}+B C^{2}=A C^{2}AB
2
+BC
2
=AC
2
\sqrt{A B^{2}+B C^{2}}=A C
AB
2
+BC
2
=AC
=\sqrt{12^{2}+5^{2}}=
12
2 +5
7
=\sqrt{144+25}=13=
144+25
=13
Now the question says to find 1) Sin A and Tan A, 2) Sin C and Cot C.
So using the formula, we get
\sin A=\frac{\text {height}}{\text {hypotenuse}}=\frac{B C}{A C}=\frac{5}{13}sinA=
hypotenuse
height = AC
BC = 135
\tan A=\frac{\text {height}}{\text {base}}=\frac{B C}{A B}=\frac{5}{12}tanA=base
height= AB
BC = 125
\sin C=\frac{\text {height}}{\text {hypotenuse}}=\frac{A B}{A C}=\frac{12}{13}sinC=
hypotenuse
height= AC
AB = 1312
\cot C=\frac{\text {base}}{\text {height}}=\frac{B C}{A B}=\frac{5}{13}cotC=
height
base=AB
BC=135
Therefore, the value of \sin A=\frac{5}{13}, \ {Tan} A=\frac{5}{12}, \sin C=\frac{12}{13}, \quad \cot C=\frac{5}{13}sinA=
13
5 , TanA= 12
5,sinC=
13
12 ,cotC.... ..