2Â years, 6Â months ago
2Â years, 6Â months ago
Thank you sir....i will see and post at the earliest.....
2Â years, 6Â months ago
Hi
For Question 50 even if you use abc/4A, answer should come same
In triangle MCD , CD=12 MC = MD , now MC can be calculated using cosine rule
cos 150 = (MB^2+BC^2-MC^2)/2MB.BC
So MC^2 = 288+144rt3
Now area of triangle is 1/2 *2* (12+6rt3) (height of equi triangle +side of square will be the height of triangle MCD)
NOW R = 12 *MC^2/4area
= 12 (288+144rt3)/4*0.5*12*(12+6rt3)
= 288+144rt3/24+12rt3
so R =12cm.
And for Q67 can you please post your solution so that I can check if you are making any mistake .