Find the angle of a triangle

Question

I'v tried to solve this problem but did not get the right result.

Triangel PQR is PQ = 5,0 cm, QR = 6,3 cm and RP = 7,4 cm. Calculate angle P.

I tried to solve it by using by using the following formula $c^2 = a^2 + b^2 - 2abcosP$. The result I get is 41 degrees but the correct result seems to be 57 degrees.

I used the following steps.

$cosP = (a^2+b^2-c^2)/(2 \cdot a \cdot b )$

Thanks!

By the way, why do you denote $;\cos C;$ instead of $;\cos P;$ or even $;\cos\angle QPR;$ ? Perhaps this misled you...? — Timbuc, Sep 04 '14 at 15:45
@S4M1R, it still makes no much sense in your question, as what is $;a,b,c;$ ? Do you mean $;a=PQ;$ , or perhaps $;a=PR;$ ...? You must be way more careful with your notation. — Timbuc, Sep 04 '14 at 15:52

score 2 · Accepted Answer · answered Sep 04 '14 at 15:44

2

Indeed the solution I get is $\;57.215^\circ\;$ . You have the right formula but maybe you messed between the sides. It should be

$$\cos\alpha=\frac{5^2+7.4^2-6.3^2}{74}$$

answered Sep 04 '14 at 15:44

Timbuc

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Find the angle of a triangle

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