L and M are two points 5 cm apart. N is another point such that NLM = 168° and NML = 80°. Construct a…

Question

L and M are two points 5 cm apart. N is another point such that NLM = 168° and NML = 80°. Construct a perpendicular from N to meet LM at P. Measure NP.

ScanSolve AI Tutor · Accepted Answer

Ili kutatua swali hili, tutatumia kanuni za trigonometry. Kwa kuwa haiwezekani kujenga kimwili, tutakokotoa urefu wa NP.

Kwanza, tunahitaji kuelewa pembe za pembetatu LNM.
Pembe zilizopewa ni $\angle NLM = 168^\circ$ na $\angle NML = 80^\circ$ .
Jumla ya pembe mbili hizi ni $168^\circ + 80^\circ = 248^\circ$ . Hii ni zaidi ya $180^\circ$ , ambayo inamaanisha kwamba pembe hizi haziwezi kuwa pembe za ndani za pembetatu kwa wakati mmoja.
Kwa kawaida, katika matatizo ya ujenzi, ikiwa pembe moja ni kubwa kuliko $90^\circ$ na inafanya jumla ya pembe kuwa zaidi ya $180^\circ$ , basi pembe ya ndani ya pembetatu inachukuliwa kuwa nyongeza ya pembe iliyotolewa.
Kwa hivyo, tutachukulia kwamba pembe ya ndani ya pembetatu LNM kwenye kipeo L ni nyongeza ya $168^\circ$ .

Tafuta pembe za ndani za $\triangle LNM$ .
- Pembe ya ndani kwenye L: $\angle L = 180^\circ - 168^\circ = 12^\circ$ .
- Pembe ya ndani kwenye M: $\angle M = 80^\circ$ .
- Pembe ya ndani kwenye N: $\angle N = 180^\circ - (\angle L + \angle M) = 180^\circ - (12^\circ + 80^\circ) = 180^\circ - 92^\circ = 88^\circ$ . Sasa tuna pembetatu halali LNM yenye LM = 5 cm, $\angle L = 12^\circ$ , $\angle M = 80^\circ$ , na $\angle N = 88^\circ$ .
Tumia Kanuni ya Saini (Sine Rule) kupata urefu wa upande LN.
Kanuni ya Saini inasema: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ .
Katika $\triangle LNM$ :
$\frac{LN}{\sin(\angle M)} = \frac{LM}{\sin(\angle N)}$ $\frac{LN}{\sin(80^\circ)} = \frac{5 \text{ cm}}{\sin(88^\circ)}$ $LN = \frac{5 \times \sin(80^\circ)}{\sin(88^\circ)}$ $LN \approx \frac{5 \times 0.98480775}{0.99939083}$ $LN \approx 4.9270 \text{ cm}$
Kokotoa urefu wa mstari wima NP.
Mstari wima kutoka N unakutana na LM kwenye P. Kwa kuwa $\angle L = 12^\circ$ na $\angle M = 80^\circ$ zote ni pembe kali, P iko kati ya L na M.
Fikiria pembetatu ya kulia $\triangle LNP$ , ambapo pembe kwenye P ni $90^\circ$ .
$\sin(\angle L) = \frac{\text{upande wa pili}}{\text{hypotenuse}} = \frac{NP}{LN}$ $NP = LN \sin(\angle L)$ $NP = 4.9270 \times \sin(12^\circ)$ $NP \approx 4.9270 \times 0.20791169$ $NP \approx 1.0244 \text{ cm}$
Ukizungusha hadi sehemu mbili za desimali, tunapata:
$NP \approx 1.02 \text{ cm}$
Urefu wa NP ni takriban $\text{1.02 cm}$ .

L and M are two points 5 cm apart. N is another point such that NLM = 168° and NML = 80°. Construct a perpendicular from N to meet LM at P. Measure NP.

Answer

Step-by-step solution

Tafuta pembe za ndani za $\triangle LNM$ .

Tumia Kanuni ya Saini (Sine Rule) kupata urefu wa upande LN.

Kokotoa urefu wa mstari wima NP.

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More Mathematics Questions

L and M are two points 5 cm apart. N is another point such that NLM = 168° and NML = 80°. Construct a perpendicular from N to meet LM at P. Measure NP.

Answer

Step-by-step solution

Tafuta pembe za ndani za △LNM\triangle LNM△LNM.

Tumia Kanuni ya Saini (Sine Rule) kupata urefu wa upande LN.

Kokotoa urefu wa mstari wima NP.

Try one yourself

Still stuck on this one?

More Mathematics Questions

Tafuta pembe za ndani za $\triangle LNM$ .