There is no solution and it’s true
The angles and lengths of each of the given triangles are;
5) m∠B = 57.52°
6) B = 70.81°
7) AB = 55.43 Km
8) AC = 39.06 ft
<h3>How to use cosine rule?</h3>
The cosine rule is expressed as;
c = √[a² + b² - 2ab(cos C)]
5) Using cosine rule;
BC = √[21² + 13² - 2(21*13)(cos 91)]
BC = 24.89
Using sine rule, we can find angle B as;
21/sin m∠B = 24.89/sin 91
sin m∠B = (21 * sin 91)/24.89
sin m∠B = 0.8436
m∠B = sin⁻¹0.8436
m∠B = 57.52°
6) Using cosine rule;
14² = 11² + 13² - 2(11*13)(cos B)]
196 = 121 + 169 - 286(cos B)
cos B = (121 + 169 - 196)/286
cos B = 0.3287
B = cos⁻¹0.3287
B = 70.81°
7) Using cosine rule;
AB = √[24² + 36² - 2(24*36)(cos 134)]
AB = 55.43 Km
8) Using cosine rule;
AC = √[21² + 26² - 2(21*26)(cos 112)]
AC = 39.06 ft
Read more about cosine rule at; brainly.com/question/4372174
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Answer: the no.s are-
2.472
-6.472
Step-by-step explanation:
xy=24-----1
x+y=-4------2
from 2 we get x=-4-y
sustituting=
into 1 we get
quadratic eq = y^2 + 4y + 24=0
so
-4+- root 16-96/2
thus on further solving we get -2 +-4.472
on adding= -2+4.472= 2.472
on subtraction= -2 -4.472= -6.472
Answer:
<h2>
∠PQT = 72°</h2>
Step-by-step explanation:
According to the diagram shown, ∠OPQ = ∠OQP = 18°. If PQT is a tangent to the circle, it can be inferred that line OQ is perpendicular to line QT. Ths shows that ∠OQT = 90°.
Also from the diagram, ∠OQP + ∠PQT = ∠OQT;
∠PQT = ∠OQT - ∠OQP
Given ∠OQP = 18° and ∠OQT = 90°
∠PQT = 90°-18°
∠PQT = 72°
