pls solve with solutions
Answers
Q1
cos A= b²+c²-a²/2bc
=9+16-4/24
= 21/24
Q2
a/sinA= b/sinB
2/2/3=3/sinB
2*3/2=3/sinB
3=3/sinB
1= sinB
Q3
Let the angles of triangle be a-d, a , a +
d
i.e ∠A = a-d
∠B =
a
∠C =
a+d
we know, sum of angles of a triangle = 180°
h e n c e, ∠A + ∠B + ∠C = 180 °
a-d + a + a + d = 180°
3 a = 180 °
a = 60 °
T h e r e f o r e, ∠B = 6 0°
It is given b/c=√3/√2................... (1)
from sine rule
a /sin A =b /sin B =c /sin C
a/sin A =b/c sin B/ sin C
a /sin A =√ 3 /√ 2 =sin B /sin C
sinC=√2/√3*sinB
sinC=√2/√3*√3/2
sinC=1/√2
C = 45
∠A+∠B+∠C=180°
∠A+60+45=180
∠A+105=180
∠A=75
Q4
A=√s(s﹣a)(s﹣b)(s﹣c)
s=a+b+c/2= 21
s-a = 21-13 = 8
s-b = 7
s-c = 6
area= √(21*8*7*6)
= √7056
=84
Q6 sry i can only tell u
For any triangle ABC, using cosine rule we can write bccosA = (b2+c2-a2)/2. Now
take the LCM of the given terms in the questions and replace the value of
bccosA. and simplify it. similarly do for ca*cosB and ab* cosC. You simplifying
you will get all the terms equal to(b2+c2+a2)/2.