Question
In a triangle ABC, if 5∠A = 3∠B = 15∠C, then find the
value of cosA + sin(C + 10 ° )Solution
Given, 5∠A = 3∠B = 15∠C
Or, ∠A:∠B:∠C = (1/5):(1/3):(1/15) = 3:5:1
Let ∠A, ∠B and ∠C be '3x', '5x' and 'x', respectively
Therefore, 3x + 5x + x = 180 o (sum of the interior angles of a triangle is 180 o )
Or, 9x = 180 o
Or, x = 20 o
Therefore, cosA + sin(C + 10 o ) = cos60 o + sin30 o = (1/2) + (1/2) = 1
√3598 × √(230 ) ÷ √102= ?
15% of 2400 + (√ 484 – √ 256) = ?
(13)2 - 3127 ÷ 59 = ? x 4
6269 + 0.25 × 444 + 0.8 × 200 = ? × 15
...(53 + 480 ÷ 4)% of 20 = ?% of 70
Find the simplified value of the following expression:
62 + 122 × 5 - {272 + 162 - 422}
(15 × 225) ÷ (45 × 5) + 480 = ? + 25% of 1240
√ [? x 11 + (√ 1296)] = 16
11 × 25 + 12 × 15 + 14 × 20 + 15 = ?