Hence find the length of MN and AC

Question

ScanSolve AI Tutor · Accepted Answer

Ili kupata urefu wa MN na AC, tutatumia Kanuni ya Kosine (Law of Cosines) kwa kila pembetatu.

Kwa pembetatu LMN:
Tunapewa:

Tumia Kanuni ya Kosine kupata urefu wa MN.
Kanuni ya Kosine inasema $c^2 = a^2 + b^2 - 2ab\cos(C)$ . Katika kesi hii, $c = MN$ , $a = LM$ , $b = LN$ , na $C = \angle L$ .
$MN^2 = LM^2 + LN^2 - 2(LM)(LN)\cos(\angle L)$ $MN^2 = (2\, \text{cm})^2 + (4\, \text{cm})^2 - 2(2\, \text{cm})(4\, \text{cm})\cos(50^\circ)$
Kokotoa thamani ya $MN^2$ .
$MN^2 = 4\, \text{cm}^2 + 16\, \text{cm}^2 - 16\, \text{cm}^2 \times \cos(50^\circ)$ $MN^2 = 20\, \text{cm}^2 - 16\, \text{cm}^2 \times 0.6428$ $MN^2 = 20\, \text{cm}^2 - 10.2848\, \text{cm}^2$ $MN^2 = 9.7152\, \text{cm}^2$
Pata mzizi wa mraba wa $MN^2$ ili kupata MN.
$MN = \sqrt{9.7152\, \text{cm}^2}$ $MN \approx \boxed{\text{3.12 cm}}$
Kwa pembetatu ABC:
Tunapewa:
- Urefu wa AB = 5 cm
- Urefu wa BC = 7.5 cm
- Pembe B = $75^\circ$
Tumia Kanuni ya Kosine kupata urefu wa AC.
Katika kesi hii, $c = AC$ , $a = AB$ , $b = BC$ , na $C = \angle B$ .
$AC^2 = AB^2 + BC^2 - 2(AB)(BC)\cos(\angle B)$ $AC^2 = (5\, \text{cm})^2 + (7.5\, \text{cm})^2 - 2(5\, \text{cm})(7.5\, \text{cm})\cos(75^\circ)$
Kokotoa thamani ya $AC^2$ .
$AC^2 = 25\, \text{cm}^2 + 56.25\, \text{cm}^2 - 75\, \text{cm}^2 \times \cos(75^\circ)$ $AC^2 = 81.25\, \text{cm}^2 - 75\, \text{cm}^2 \times 0.2588$ $AC^2 = 81.25\, \text{cm}^2 - 19.41\, \text{cm}^2$ $AC^2 = 61.84\, \text{cm}^2$
Pata mzizi wa mraba wa $AC^2$ ili kupata AC.
$AC = \sqrt{61.84\, \text{cm}^2}$ $AC \approx \boxed{\text{7.86 cm}}$

Step-by-step solution