[amp_mcq option1=”sin c sin B” option2=”cos c cos B” option3=”tan c tan B” option4=”sin c cos B” correct=”option3″]<\/p>\n
The correct answer is $\\boxed{\\text{B. cos c cos B}}$.<\/p>\n
In a spherical triangle ABC right angled at C, the sine of the angle B is equal to the cosine of the angle C times the cosine of the angle B. This can be shown using the law of cosines for spherical triangles, which states that<\/p>\n
$$c^2 = a^2 + b^2 – 2ab \\cos C$$<\/p>\n
where $c$ is the side opposite the right angle, $a$ and $b$ are the other two sides, and $C$ is the angle opposite side $c$.<\/p>\n
In this case, we know that $C = 90^\\circ$, so the law of cosines becomes<\/p>\n
$$c^2 = a^2 + b^2 – 2ab$$<\/p>\n
We can then take the sine of both sides to get<\/p>\n
$$\\sin c = \\frac{\\sin a}{\\cos B}$$<\/p>\n
or<\/p>\n
$$\\sin B = \\cos c \\cos B$$<\/p>\n
Therefore, the sine of the angle B is equal to the cosine of the angle C times the cosine of the angle B.<\/p>\n
The other options are incorrect because they do not take into account the fact that the triangle is right angled at C. For example, option A would be correct if the triangle were not right angled, but in this case it is not.<\/p>\n","protected":false},"excerpt":{"rendered":"
[amp_mcq option1=”sin c sin B” option2=”cos c cos B” option3=”tan c tan B” option4=”sin c cos B” correct=”option3″]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[654],"tags":[],"class_list":["post-8150","post","type-post","status-publish","format-standard","hentry","category-surveying","no-featured-image-padding"],"yoast_head":"\n