In a right angle Triangle ABC right angle is at c.BC +CA =23 cm and BC-CA=7cm then find sin A and tan B
Answers
BC + AC= 23 ----( i ) BC - AC = 7 ----( ii ) From eqn (i), BC = 23 - AC Substituting the value of BC in eqn(ii), => 23 -AC -AC =7 => -2AC= 7-23 = -16 => AC= 8cm Putting the value of AC in eqn (i), => BC+ 8 =23 => BC=23-8=15 => BC= 15 cm Thus, p = 8cm b = 15 cm h = root over of (15)^2 + (8)^2 = root over of 225 + 64 = root 289 = 17 cm Sin A = BC/AC= 15/17 Tan B= AC/BC= 8/17 Hope it may help you ..
Given,
ABC is a right angle triangle, right angle is at C.
and BC+CA = 23cm, BC-CA = 7cm.
Then, SinA and TanB = ?
From both BC+CA = 23cm and BC-CA = 7cm.
BC = 23-CA ............... Equation (i)
BC = 7+CA ............... Equation (ii)
Now add both the equations.
BC = 23-CA
BC = 7+CA
.....................
2BC = 30
BC = 30/2
BC = 15CM
Now substitute the value of BC in equation i or ii.
BC = 7+CA
15 = 7+CA
15-7 = CA
8 = CA
OR
AC = 8CM.
By pythagoras theorem,
(AB)^2 = (AC)^2+(BC)^2
(AB)^2 = 8^2+15^2
(AB)^2 = 64+225
(AB)^2 = 289
AB = Square root of 289
AB = 17cm.
SinA = opposite side to A/hypotenuse
SinA = 15/17
TanB =opposite side to B/adjacent side to B.
Tan B = 8/15
