If angle JKM= 15x-48 and angle L= 5x+12 and angle M= 40, find angle MKL
![If angle JKM 15x48 and angle L 5x12 and angle M 40 find angle MKL class=](https://us-static.z-dn.net/files/d95/d2c9773897afe5f683734754834cd1b6.jpg)
The required angle MKL = 96⁰
In a Euclidean space, the sum of angles of a triangle equals the straight angle. A triangle has three angles, one at each vortex, bounded by a pair of adjacent sides.
Mathematically,if ABC is a traingle with sides A,B and C. Then,
∠A + ∠B + ∠C = 180⁰
Now,it is given that
∠JKM = 15x-48
∠L= 5x+12
∠M= 40
So, In ΔMKL,
Since sum of all the angles = 180⁰
⇒ ∠K + ∠L + ∠M = 180⁰
⇒ ∠JKM + (5x+12) + 40 = 180⁰
⇒ (15x-48) + (5x+12) + 40 = 180⁰
⇒ 15x + 5x + 12 - 48 + 40 = 180⁰
⇒ 20x + 12 - 8 = 180⁰
⇒ 20x + 4 = 180⁰
⇒ 20x = 180⁰ - 4
⇒ 20x = 176⁰
⇒ x = 176/20⁰
⇒ x = 88/10⁰
Putting the value of x in ∠JKM. We get,
∠ JKM = 15 (88/10) - 48
⇒ ∠JKM= 132 - 48
⇒ ∠JKM= 84
∵ Angle of straight line = 180⁰
∴Taking the straight line JKL from the figure shown. We have
∠JKM + ∠MKL = 180⁰
Putting the values of ∠JKM. We have
⇒ 84 + ∠MKL = 180⁰
⇒ ∠MKL = 180⁰ - 84
⇒ ∠MKL = 96⁰
Hence, the required angle MKL is 96⁰.
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